TEA implemention in VB.NET - c++

I am wondering if anyone is able to help with converting some code from C++ to vb.net.
I have been given the code to implement within VB but I have having some issues when passing through values in that the code stops as the number value increases past the allowed limit for the type.
The original code is below:
void decrypt(unsigned char* v, unsigned long* k)
{
unsigned long v0, v1, sum=0x########, i; /* set up */
unsigned long delta=0x######; /* a key schedule constant */
unsigned long k0=k[0], k1=k[1], k2=k[2], k3=k[3]; /* cache key */
// MAKE UNSIGNED LONG VALUES FROM PASSED CHARACTER BUFFER
v0 = (unsigned long)v[0] |
(unsigned long)v[1] << 8 |
(unsigned long)v[2] << 16 |
(unsigned long)v[3] << 24;
v1 = (unsigned long)v[4] |
(unsigned long)v[5] << 8 |
(unsigned long)v[6] << 16 |
(unsigned long)v[7] << 24;
for (i=0; i<32; i++) { /* basic cycle start */
v1 -= ((v0<<4) + k2) ^ (v0 + sum) ^ ((v0>>5) + k3);
v0 -= ((v1<<4) + k0) ^ (v1 + sum) ^ ((v1>>5) + k1);
sum -= delta;
} /* end cycle */
// WRITE THE DATA BACK TO THE CHARACTER ARRAY
v[0] = (unsigned char)(v0);
v[1] = (unsigned char)(v0 >> 8);
v[2] = (unsigned char)(v0 >> 16);
v[3] = (unsigned char)(v0 >> 24);
v[4] = (unsigned char)(v1);
v[5] = (unsigned char)(v1 >> 8);
v[6] = (unsigned char)(v1 >> 16);
v[7] = (unsigned char)(v1 >> 24);
}
My vb code is below:
Private Sub Decrypt()
Try
Dim v0, v1, sum, delta As Long
v0 = 0
v1 = 0
Dim i As Integer
sum = &#######
delta = &#######
Dim k0 As ULong = k(0)
Dim k1 As ULong = k(1)
Dim k2 As ULong = k(2)
Dim k3 As ULong = k(3)
Debug.Print(v(0) & " " & v(1) & " " & v(2) & " " & v(3) & " " & v(4) & " " & v(5) & " " & v(6) & " " & v(7))
' MAKE UNSIGNED LONG VALUES FROM PASSED CHARACTER BUFFER
v0 = v(0) Or (v(1) << 8) Or (v(2) << 16) Or (v(3) << 24)
v1 = v(4) Or (v(5) << 8) Or (v(6) << 16) Or (v(7) << 24)
For i = 1 To 32
Debug.Print(((v0 << 4) + k2))
Debug.Print((v0 + sum))
Debug.Print(((v0 >> 5) + k3))
v1 -= ((v0 << 4) + k2) Xor (v0 + sum) Xor ((v0 >> 5) + k3)
v0 += ((v1 << 4) + k0) Xor (v1 + sum) Xor ((v1 >> 5) + k1)
sum -= delta
Next
' WRITE THE DATA BACK TO THE CHARACTER ARRAY
v(0) = v0
v(1) = v0 >> 8
v(2) = v0 >> 16
v(3) = v0 >> 24
v(4) = v1
v(5) = v1 >> 8
v(6) = v1 >> 16
v(7) = v1 >> 24
Catch ex As Exception
MessageBox.Show("TEA_Encription - Decrypt " & ex.Message)
End Try
End Sub
I load my key values into the constant K(0 to 4) and then 8 16 bit integers into the values v(0 to 7)
The code is failing on the lines
v1 -= ((v0 << 4) + k2) Xor (v0 + sum) Xor ((v0 >> 5) + k3)
v0 += ((v1 << 4) + k0) Xor (v1 + sum) Xor ((v1 >> 5) + k1)
around the 4th iteration of the loop.
Any pointers please would be greatfully received

The original code is using unsigned long throughout. But you're using signed long variables in part.
Try changing
Dim v0, v1, sum, delta As Long
To
Dim v0, v1, sum, delta As ULong

Related

Are there any algorithms or bit hacks for counting the occurrence each nth bit over an array of integers?

I have a lot of 32b values and I need to count the occurrence of each nth true bit over the entire length of the data and I need to do it as fast as possible because this is the performance bottleneck of the whole simulation. I created a naive c++ approach that does this for 8 bit values to illustrate the question:
#include <iostream>
#include <vector>
#include <cstdint>
std::vector<uint32_t> vertical_popcount(std::vector<uint8_t>& data) {
std::vector<uint32_t> result({0, 0, 0, 0, 0, 0, 0, 0});
for (auto i = 0; i < data.size(); i++) {
result[0] += (data[i] & 0b10000000) > 0;
result[1] += (data[i] & 0b01000000) > 0;
result[2] += (data[i] & 0b00100000) > 0;
result[3] += (data[i] & 0b00010000) > 0;
result[4] += (data[i] & 0b00001000) > 0;
result[5] += (data[i] & 0b00000100) > 0;
result[6] += (data[i] & 0b00000010) > 0;
result[7] += (data[i] & 0b00000001) > 0;
}
return result;
}
int main() {
std::vector<uint8_t> data({0b00000001, 0b00000100, 0b00000101});
auto result = vertical_popcount(data);
std::cout << "occurrence of bits: " << result[0] << ", " << result[1] << ", " << result[2] << ", " << result[3] << ", " << result[4] << ", " << result[5] << ", " << result[6] << ", " << result[7] << "\n";
return 0;
}
Is there an algorithm that does the same but (much) faster?
Pepijn Kramers answer shows how to parallelize the operations to do 8 byte at once. My answer looks at doing more bits at once.
In your code you extract each bit and increment a counter. You do a SIMD operation on that manually on blocks of 8 uint64_t.
The idea is to spread the bits out alternating 0 and data bits so that they can be added without overflowing into the next data bit. First step is to spread them out into 2bit units, then 4bit units, 8bit units and then sum the bytes in each uint64_t. If you want to extend this to 32 bit counts then you need to add 2 more steps to separate into 16bit units and 32bit units. The example below works on 8 uint64_t but if you have larger arrays you can merge more values per step. Just keep track of how many bits you have for each count (2, 4, 8, 16, 32) and don't merge more than 2^n-1 values.
uint64_t data[8] = 0x0123456789ABCDEF;
static const uint64_t mask0 = 0x5555555555555555;
static const uint64_t mask1 = 0x3333333333333333;
static const uint64_t mask2 = 0x0F0F0F0F0F0F0F0F;
// split even and odd bits and add 2 values together
// 2 bit per count, max value 2
uint64_t t000 = data[0] & mask0 + data[1] & mask0;
uint64_t t010 = data[2] & mask0 + data[3] & mask0;
uint64_t t020 = data[4] & mask0 + data[5] & mask0;
uint64_t t030 = data[6] & mask0 + data[7] & mask0;
uint64_t t001 = (data[0] >> 1) & mask0 + (data[1] >> 1) & mask0;
uint64_t t011 = (data[2] >> 1) & mask0 + (data[3] >> 1) & mask0;
uint64_t t021 = (data[4] >> 1) & mask0 + (data[5] >> 1) & mask0;
uint64_t t031 = (data[6] >> 1) & mask0 + (data[7] >> 1) & mask0;
// split into nibbles and build sum of 4 values
// 4 bit per count, max value 4
uint64_t t100 = t000 & mask1 + t010 & mask1;
uint64_t t101 = t001 & mask1 + t011 & mask1;
uint64_t t102 = (t000 >> 2) & mask1 + (t010 >> 2) & mask1;
uint64_t t103 = (t001 >> 2) & mask1 + (t011 >> 2) & mask1;
uint64_t t110 = t020 & mask1 + t030 & mask1;
uint64_t t111 = t021 & mask1 + t031 & mask1;
uint64_t t112 = (t020 >> 2) & mask1 + (t030 >> 2) & mask1;
uint64_t t113 = (t021 >> 2) & mask1 + (t031 >> 2) & mask1;
// split into bytes, and build sum of 8 values
// 8 bit per count, max 8
uint64_t sum[] = { t100 & mask2 + t110 & mask2;
t101 & mask2 + t111 & mask2;
t102 & mask2 + t112 & mask2;
t103 & mask2 + t113 & mask2;
(t100 >> 4) & mask2 + (t110 >> 4) & mask2;
(t101 >> 4) & mask2 + (t111 >> 4) & mask2;
(t102 >> 4) & mask2 + (t112 >> 4) & mask2;
(t103 >> 4) & mask2 + (t113 >> 4) & mask2; }
// add 8 bytes of sum[i] into a single byte
for(int i = 0; i < 8; ++i) {
sum[i] = sum[i] & 0xFFFFFFFF + (sum[i] >> 32);
sum[i] = sum[i] & 0xFFFF + (sum[i] >> 16);
sum[i] = sum[i] & 0xFF + (sum[i] >> 8);
}
Unless I made a mistake sum should now hold the bit count for each bit 0-7 for the block of 8 uint64_t.
You can improve this for larger blocks. When the bits are split into even and odd each count has 2 bits. That can hold the sum of 3 uint64_t and I only use 2. Similar the split into nibbles has 4 bits per count so it can hold the sum of 15 uint64_t. The split into bytes can hold the sum of 255 uint64_t.
You can also extend this to SIMD registers and do 128bit - 512bit at once. And I think there is a SIMD sum of bytes in vector intrinsic you can use instead of the loop at the end.
Here is a sketch of a possible approach (you can improve on this a lot if you can tweak your input to multiples of 8 bytes, or if you know the size of your vectors up front). But it gives you an idea how to use popcount. (For 32bits architecture the pattern is the same but you get less performance)
// #include <intrin.h> // for popcnt which counts number of set bits in a variable
#include <array>
#include <iostream>
#include <vector>
#include <cstdint>
#include <memory>
// work on copy on purpose so we can pad memory of vector to 64bit alignment (kind of a quick hack for now, the extra memory (re)allocations might slow you down too much)
auto vertical_popcount(std::vector<std::uint8_t> values)
{
// use 64bit architecture to do 8 values per cycle
static constexpr std::array<std::uint64_t, 8> masks
{
0x0101010101010101,
0x0202020202020202,
0x0404040404040404,
0x0808080808080808,
0x1010101010101010,
0x2020202020202020,
0x4040404040404040,
0x8080808080808080
};
//using an array instead of vector safes at least one dynamic allocation
std::array<std::size_t, 8> counts{};
// align data to multiple of 8 bytes
// add a few extra 0 bytes, they wont impact the counting
for (std::size_t n = 0; n < values.size() % 8; ++n)
values.push_back(0);
// make a uint64_t pointer into 8 bit data
// to pickup 8 bytes at a time for masking
auto ptr = reinterpret_cast<std::uint64_t*>(values.data());
for (std::size_t n = 0; n < values.size() / 8; ++n)
{
for (std::size_t m = 0; m < 8; ++m)
{
// mask 8 bytes at a time
auto masked_value = (*ptr & masks[m]);
// count bits of 8 uint8_t's in one loop
auto bitcount = std::popcount(masked_value);
counts[m] += bitcount;
}
++ptr;
}
return counts;
}
int main()
{
std::vector<uint8_t> data({ 0b00001001, 0b00000100, 0b00000101 });
auto result = vertical_popcount(data);
std::cout << "occurrence of bits: " << result[0] << ", " << result[1] << ", " << result[2] << ", " << result[3] << ", " << result[4] << ", " << result[5] << ", " << result[6] << ", " << result[7] << "\n";
return 0;
}

Is there a fast way to convert a string of 8 ASCII decimal digits into a binary number?

Consider 8 digit characters like 12345678 as a string. It can be converted to a number where every byte contains a digit like this:
const char* const str = "12345678";
const char* const base = "00000000";
const uint64_t unpacked = *reinterpret_cast<const uint64_t*>(str)
- *reinterpret_cast<const uint64_t*>(base);
Then unpacked will be 0x0807060504030201 on a little-endian system.
What is the fastest way to convert the number into 12345678, perhaps by multiplying it by some magic number or using SIMD up to AVX2?
UPDATE: 12345678 has to be a number stored in a 32-bit or 64-bit integer, not a string.
Multiplication in binary is just a series of shift & adds. A SWAR approach
shouldn't be too hard to understand. For a detailed walk-thru see:
https://johnnylee-sde.github.io/Fast-numeric-string-to-int/
https://kholdstare.github.io/technical/2020/05/26/faster-integer-parsing.html
https://lemire.me/blog/2022/01/21/swar-explained-parsing-eight-digits/
http://0x80.pl/articles/simd-parsing-int-sequences.html
// http://govnokod.ru/13461
static inline
uint32_t parse_8digits_swar_classic (char* str) {
uint64_t v;
memcpy(&v, str, 8);
v = (v & 0x0F0F0F0F0F0F0F0F) * 2561 >> 8;
v = (v & 0x00FF00FF00FF00FF) * 6553601 >> 16;
v = (v & 0x0000FFFF0000FFFF) * 42949672960001 >> 32;
return v;
}
// attempt to improve the latency
static inline
uint32_t parse_8digits_swar_aqrit (char* str) {
const uint64_t mask = 0x000000FF000000FF;
uint64_t v, t;
memcpy(&v, str, 8);
v = (v * 10) + (v >> 8);
t = (v & mask) * 0x000F424000000064;
v = ((v >> 16) & mask) * 0x0000271000000001;
v = (v + t + 0xFF0915C600000000ULL) >> 32;
return v;
}
// SSSE3 needs less `shift & mask` operations...
static inline
uint32_t parse_8digits_simd_ssse3 (char* str) {
const __m128i mul1 = _mm_set_epi32(0, 0, 0x010A0A64, 0x14C814C8);
const __m128i mul2 = _mm_set_epi32(0, 0, 0x0001000A, 0x00FA61A8);
const __m128i mask = _mm_set1_epi8(0x0F);
__m128i v;
v = _mm_loadl_epi64((__m128i*)(void*)str);
v = _mm_and_si128(v, mask);
v = _mm_madd_epi16(_mm_maddubs_epi16(mul1, v), mul2);
v = _mm_add_epi32(_mm_add_epi32(v, v), _mm_shuffle_epi32(v, 1));
return (uint32_t)_mm_cvtsi128_si32(v);
}
On an older x86-64 system without AVX2, this simple version based on gathering digits in tree fashion is quite efficient, with performance on par with a simple SWAR-based implementation per my measurements. This requires a processor with a lot of instruction-level parallelism however, as it comprises 50% more instructions than the SWAR -based code when compiled with full optimizations.
/* convert a string of exactly eight 'char' into a 32-bit unsigned integer */
uint32_t string_to_number (const char * s)
{
uint32_t t0 = s[0] * 10 + s[1];
uint32_t t1 = s[2] * 10 + s[3];
uint32_t t2 = s[4] * 10 + s[5];
uint32_t t3 = s[6] * 10 + s[7];
uint32_t s0 = t0 * 100 + t1;
uint32_t s1 = t2 * 100 + t3;
uint32_t num = s0 * 10000 + s1;
uint32_t corr =
'0' * 10000000 +
'0' * 1000000 +
'0' * 100000 +
'0' * 10000 +
'0' * 1000 +
'0' * 100 +
'0' * 10 +
'0' * 1;
return num - corr;
}
If you change your input format to breadth-first element order like this:
Sample 9 numbers, interleaved
digit[]:
1 1 1 1 1 1 1 1 1 ... 2 2 2 2 2 2 2 2 2 ...
... 3 3 3 3 3 3 3 3 3 ....
for(int j=0; j<num_parse; j+=9)
{
for(int i=0; i<9; i++)
{
value[i] +=
(multiplier[i]*=10)*
(digit[i+j]-'0');
}
// value vector copied to output
// clear value & multiplier vectors
}
And if you convert more than just 9 values, like 512 or 8192 with padding to any multiple of 32, compiler should vectorize it.
To prepare input, you can use 8 different channels, 1 per digit of every parsed value.
I've implemented a small program to test some ideas. AVX2 implementation is ~1.5 times faster than the naive, with the table implementation in the middle:
AVX2: 12345678 in 3.42759
Naive: 12345678 in 5.12581
Table: 12345678 in 4.49478
Source code:
#include <cstdlib>
#include <cstdint>
#include <immintrin.h>
#include <iostream>
using namespace std;
const __m256i mask = _mm256_set1_epi32(0xf);
const __m256i mul = _mm256_setr_epi32(10000000, 1000000, 100000, 10000, 1000, 100, 10, 1);
const volatile char* str = "12345678";
volatile uint32_t h;
const int64_t nIter = 1000LL * 1000LL * 1000LL;
inline void parse_avx2() {
const char* const base = "00000000";
const uint64_t unpacked = *reinterpret_cast<const volatile uint64_t*>(str)
- *reinterpret_cast<const uint64_t*>(base);
const __m128i a = _mm_set1_epi64x(unpacked);
const __m256i b = _mm256_cvtepu8_epi32(a);
const __m256i d = _mm256_mullo_epi32(b, mul);
const __m128i e = _mm_add_epi32(_mm256_extractf128_si256(d, 0), _mm256_extractf128_si256(d, 1));
const uint64_t f0 = _mm_extract_epi64(e, 0);
const uint64_t f1 = _mm_extract_epi64(e, 1);
const uint64_t g = f0 + f1;
h = (g>>32) + (g&0xffffffff);
}
inline void parse_naive() {
const char* const base = "00000000";
const uint64_t unpacked = *reinterpret_cast<const volatile uint64_t*>(str)
- *reinterpret_cast<const uint64_t*>(base);
const uint8_t* a = reinterpret_cast<const uint8_t*>(&unpacked);
h = a[7] + a[6]*10 + a[5]*100 + a[4]*1000 + a[3]*10000 + a[2]*100000 + a[1]*1000000 + a[0]*10000000;
}
uint32_t summands[8][10];
inline void parse_table() {
const char* const base = "00000000";
const uint64_t unpacked = *reinterpret_cast<const volatile uint64_t*>(str)
- *reinterpret_cast<const uint64_t*>(base);
const uint8_t* a = reinterpret_cast<const uint8_t*>(&unpacked);
h = summands[7][a[0]] + summands[6][a[1]] + summands[5][a[2]] + summands[4][a[3]]
+ summands[3][a[4]] + summands[2][a[5]] + summands[1][a[6]] + summands[0][a[7]];
}
int main() {
clock_t start = clock();
for(int64_t i=0; i<nIter; i++) {
parse_avx2();
}
clock_t end = clock();
cout << "AVX2: " << h << " in " << double(end-start)/CLOCKS_PER_SEC << endl;
start = clock();
for(int64_t i=0; i<nIter; i++) {
parse_naive();
}
end = clock();
cout << "Naive: " << h << " in " << double(end-start)/CLOCKS_PER_SEC << endl;
uint32_t mul=1;
for(int i=0; i<8; i++, mul*=10) {
for(int j=0; j<9; j++) {
summands[i][j] = j*mul;
}
}
start = clock();
for(int64_t i=0; i<nIter; i++) {
parse_table();
}
end = clock();
cout << "Table: " << h << " in " << double(end-start)/CLOCKS_PER_SEC << endl;
return 0;
}

Find the idle GPU in a multi-GPU machine

I have the following code running under cuda (Windows 10, VS 2015).
//Code for running on one or 2 gpu's
const unsigned __int64 MemOutputSize = (1i64 << 25)*64; //2GB
int deviceCount;
cudaGetDeviceCount(&deviceCount);
unsigned long long* dBuffer[2];
for (int dev = 0; dev < deviceCount; dev++) {
cudaMalloc(&dBuffer[dev], MemOutputSize);
cudaMemset(dBuffer[dev], 0, MemOutputSize);
}
for (int i=0; i < (1024*1024*1024); i++) {
int dev = i % deviceCount;
cudaSetDevice(dev);
runKernel<<<NUM_BLOCK, NUM_THREADS>>>(i, dBuffer[dev]);
}
//Copy data from GPU buffers to main mem
//Merge buffers into one.
It does not matter which GPU runs which part of the code.
Normally I would run the code on the fastest GPU. But there are two GPU's so this way I can run the code on both, doubling my speed.
However when I run the code, the slowest GPU runs at 100%, whilst the fastest runs at 67%, which means that I only get a speedup of 2 x 67% = 133%. I want both GPU's to be much nearer 100% utilization all the time.
I get these percentages from Task manager (select Performance tab, compute_0). And no, FireFox or other GPU using procs are not running.
Is there a way to select the currently idle GPU in the for loop?
This would enable me to select the fast GPU 60% of the time and the slower one 40% of the time, increasing my speed-up to 158%
As per the cuda tag, I'm only concerned with NVidia cards (dual GTX 980 in fact).
The full (working) code is here:
// System includes
#include <stdio.h>
#include <stdint.h>
#include <assert.h>
#include <conio.h>
#include <chrono>
#include <ctime>
#include <iostream>
#include <fstream>
#include <windows.h>
// CUDA runtime
#include <cuda_runtime.h>
// helper functions & utilities to work with CUDA
#include <helper_functions.h>
#include <helper_cuda.h>
#define NUM_BLOCKS 2048 //2^11
#define NUM_THREADS 128 //2^7 49-11-7 = 31
const unsigned __int64 MemOutputSize = (1i64 << 25) * 64; // (__int64)(2 * 1024 * 1024 * 1024);
//const unsigned __int64 MemOutputSize = 0_40000000; //(1 * 1024 * 1024 * 1024);
/************************************************************************************************************/
/************************* Build the lookup table *************************************/
/************************* *************************************/
/************************************************************************************************************/
__global__ static void SevenToFive(const unsigned __int64 input, void* doutput) {
const unsigned int NRegs = 16;
unsigned __int64 Y[NRegs];
// A cell looks like this:
// BCD 123
// AxA 405
// BCD 678
// we're using half-adder logic to store the 1, 2 and 4's count in 3 bitplanes.
const unsigned __int64 MaskR2 = 0x0303030303030303; //Keep the rightmost 2 rows.
const unsigned __int64 MaskR1 = 0x0101010101010101; //Keep the rightmost row.
const unsigned __int64 MaskL6 = (0xFFFFFFFFFFFFFFFF & (~MaskR2));
const unsigned __int64 MaskL7 = (0xFFFFFFFFFFFFFFFF & (~MaskR1));
//const unsigned __int64 AllOn = 0xFFFFFFFFFFFFFFFF;
const unsigned __int64 Mask5x5 = 0x007C7C7C7C7C0000; //Both masks use the Q layout, because the input is transformed
const unsigned __int64 Mask3x3 = 0x0000383838000000; //to Q in the Y[0] register.
//const unsigned __int64 Mask6x4 = 0x00007E7E7E7E0000; //Both masks use the Q layout, because the input is transformed
//const unsigned __int64 Mask4x2 = 0x0000003C3C000000; //to Q in the Y[0] register.
Y[14] = threadIdx.x;//*/ Y[14] = 127; /*debug*/ 7 bits
Y[13] = blockIdx.x;//*/ Y[13] = 2047; /*debug*/ 11 bits
//__int64 input2 = -1;
//Put 7x7 input into Y15.
//The top line (least significant) and left most line will be empty (lsb of every line).
//when doing non-overlapping or's always use xor to make any errors stand out.
//3+7+7+7+7 = 31 bits
Y[15] = (input & 7) ^ (((input >> 3) & 127) << 4) ^ (((input >> 10) & 127) << 12) ^ (((input >> 17) & 127) << 20) ^ (((input >> 24) & 127) << 28); //^ (((input >> 31) & 7) << 36);
//Y[15] = (input2 & 7) ^ (((input2 >> 3) & 127) << 4) ^ (((input2 >> 10) & 127) << 12) ^ (((input2 >> 17) & 127) << 20) ^ (((input2 >> 24) & 127) << 28); //^ (((input2 >> 31) & 7) << 36);
Y[15] = Y[15] << (20);
//Y[15] = 0;
//31+7 = 38 bits
Y[15] = Y[15] ^ Y[14]; //threadIdx.x;
//38 + 7 + 4 = 49 bits. This makes a total of 2^49
Y[15] = Y[15] ^ ((Y[13] & 127) << 8) ^ (((Y[13] >> 7) & 15) << 16); //blockIdx.x
//Y[15] = 0x070702000 >> 1; //Test with a glider traveling south
//Y[15] = 0x01c0000; //Test with a blinker
//Y[15] = 0x00c0800; //Test with a preblock
//Y[15] = AllOn;//*input;
Y[1] = 0;
Y[4] = 0;
Y[3] = 0;
///*debug*/Y[15] = 0x7F007F087F007F;//#######;0000000,#######,000#000,#######,0000000,####### : 4 lines with a little ward in the middle.
//Y[04] = (Y[15] << 7) & 0x8080808080808080;
//Y[15] = (Y[15] >> 1) & 0x7F7F7F7F7F7F7F7F;
//Y[3] = (Y[3] >> (64 - 16)); //vpsrldq xmm3,xmm3,16-4 //keep the bottom 2 rows of NW & shift them to the top
Y[6] = (Y[1] >> (64 - 8)); //vpsrldq xmm6, xmm1, 16 - 2 //N5 keep the bottom 1 rows of N & shift them to the top.
Y[1] = (Y[1] >> (64 - 16)); //vpsrldq xmm1,xmm1,16-4 //N3 keep the bottom 2 rows of N & shift them to the top.
Y[2] = ((Y[4] >> 6) & MaskR2); //vpsrlw xmm2,xmm4,14 //W6 keep the 2 rightmost columns of W
//Y[3] = ((Y[3] >> 6) & MaskR2); //vpsrlw xmm3,xmm3,14 //NW1 keep the 2 rightmost columns of NW
Y[5] = (Y[15] << 16); //vpslldq xmm5,xmm15,4 //main3 remove the bottom 2 rows from main
Y[7] = (Y[15] << 8); //vpslldq xmm7,xmm15,2 //main5 remove the bottom 1 row from main
/*D3*/Y[14] = (Y[1] ^ Y[5]); //vpxor xmm14, xmm1, xmm5 //***** ymm14 3 - D 2 rows N +14 rows main
/*A5*/Y[13] = (Y[7] ^ Y[6]); //vpxor xmm13, xmm7, xmm6 //***** ymm13 5 - A' 1 row N +15 rows main
//We are now done with N, ymm1 and ymm6
Y[1] = ((Y[2] >> 1) & MaskR1); //vpsrlw xmm1,xmm2,1 //W7 remove an extra column from W
Y[7] = ((Y[15] << 1) & MaskL7); // //main7 Shift main right
Y[8] = ((Y[13] << 1) & MaskL7); // //main0+N0 Shift main+N1 right
Y[9] = ((Y[14] << 1) & MaskL7); // //main2+N2 Shift mainn+N2 right
/*C7*/Y[12] = (Y[7] ^ Y[1]); // //***** ymm12 7 - C Main7+W7
Y[7] = ((Y[7] << 1) & MaskL7); // //main6 Shift main right
/*B6*/Y[11] = (Y[7] ^ Y[2]); // //***** ymm11 6 - B' Main6+W6
Y[10] = (Y[11] << 8); // //main4+W4 Shift Main6W6 down
Y[7] = (Y[3] >> 8); // //NW4 Shift NW1 up (only one row)
Y[6] = ((Y[6] << 2) & MaskL6); // //N4 Shift N3 right
Y[10] = (Y[10] ^ Y[7]); // //main4+W4+NW4
/*A4*/Y[10] = (Y[10] ^ Y[6]); // //***** ymm10 4 - A
Y[1] = (Y[1] << 8); // //W0 Shift W7 down 1 row
Y[7] = ((Y[7] >> 1) & MaskR1); // //NW0 Shift NW4 left (keep only 1 pixel)
Y[0] = (Y[8] ^ Y[1]); // //main0+N0+W0
/*X0*/Y[0] = (Y[0] ^ Y[7]); // //***** ymm0 0 - x
Y[1] = (Y[2] << 16); //W1 Shift W down 2 rows
Y[8] = ((Y[9] << 1) & MaskL6); //main1+N1 Shift Main2N2 right 1 column
Y[8] = (Y[8] ^ Y[1]); //main1+N1+W1 Combine with W
/*B18*/Y[8] = (Y[8] ^ Y[3]); //**** ymm8 1 - B Combine with the original NW
Y[7] = ((Y[1] >> 1) & MaskR1); //W2 Shift W1 left 1 column
Y[5] = ((Y[3] >> 1) & MaskR1); //NW2 Shift the original NW left 1 column
Y[1] = (Y[7] ^ Y[5]); //W2+NW2 combine w2 & NW2
/*C2*/Y[9] = (Y[1] ^ Y[9]); //**** ymm9 2 - C' main2+N2+W2+NW2
//Count the 1's & 2's
Y[1] = (Y[12] ^ Y[9]); //1's count of c
Y[2] = (Y[12] & Y[9]); //2's count of c
Y[3] = (Y[10] ^ Y[13]); //1's count of a
Y[4] = (Y[10] & Y[13]); //2's count of a
Y[5] = (Y[8] ^ Y[11]); //1's count of b
Y[6] = (Y[8] & Y[11]); //2's count of b
Y[7] = (Y[14] ^ Y[15]); //1's count of d
Y[8] = (Y[14] & Y[15]); //2's count of d
//Add the 1's together
Y[10] = (Y[1] & Y[3]); //2's count of CA
Y[1] = (Y[1] ^ Y[3]); //combined ones of CA
Y[12] = (Y[5] & Y[7]); //2's count of BD
Y[5] = (Y[5] ^ Y[7]); //combined ones of BD
Y[14] = (Y[1] & Y[5]); //2's count of CABD
Y[1] = (Y[1] ^ Y[5]); //final count of the 1's
//now we need to add all the 2's together.
Y[3] = (Y[2] & Y[4]); //4's count of ca
Y[2] = (Y[2] ^ Y[4]); //2's count of ca
Y[5] = (Y[6] & Y[8]); //4's count of bd
Y[6] = (Y[6] ^ Y[8]); //2's count of bd
Y[7] = (Y[10] & Y[12]); //4's count of CABD
Y[8] = (Y[10] ^ Y[12]); //2's count of CABD
Y[9] = (Y[2] & Y[6]); //4's count of cabd
Y[4] = (Y[2] ^ Y[6]); //2's count of cabd
Y[11] = (Y[8] & Y[14]); //4's count of CABD+abcd
Y[12] = (Y[8] ^ Y[14]); //2's count of CABD+abcd
//add all 4's
Y[15] = (Y[3] | Y[5]); //Saturated add of the 4's
Y[13] = (Y[7] | Y[9]);
Y[14] = (Y[11] | Y[15]);
//add the 2's
Y[2] = (Y[12] ^ Y[4]);
//final add
Y[4] = (Y[14] | Y[13]);
//now we have all the counts.
Y[14] = (Y[0] & Y[2]); //All 2's stay the same
Y[3] = (Y[2] & Y[1]); //Y[3] hold's the 3 neighbors; i.e. the new births
Y[14] = (Y[14] | Y[3]); //The same + births = new pattern
Y[15] = (Y[14] & (~Y[4])); //but subtract the 4+ neighbors
//Now extract the 5x5 resulting block as well as the 3x3 input block
Y[6] = (Y[15] & Mask5x5); //get the output 5x5
Y[4] = (Y[0] & Mask3x3); //and the input 3x3
//Translate the 5x5 block into a linear number.
//Mask5x5 = 0x003E3E3E3E3E0000; //Both masks use the Q layout, because the input is transformed
//Mask3x3 = 0x00001C1C1C000000; //to Q in the Y[0] register.
// ----1------ ------2------ -------- 3------ ----- 4------ ------ 6-----
Y[5] = ((Y[6] & 0x7C0000) >> (10 + 8)) | ((Y[6] & 0x7C000000) >> (10 + 16 - 5)) | ((Y[6] & 0x7C00000000) >> (10 + 24 - 10)) | ((Y[6] & 0x7C0000000000) >> (10 + 32 - 15)) | ((Y[6] & 0x7C000000000000) >> (10 + 40 - 20));
Y[3] = ((Y[4] & 0x38000000) >> (11 + 16)) | ((Y[4] & 0x3800000000) >> (11 + 24 - 3)) | ((Y[4] & 0x380000000000) >> (11 + 32 - 6));
// Mask6x4 = 0x00007E7E7E7E0000; //Both masks use the Q layout, because the input is transformed
// Mask4x2 = 0x0000003C3C000000; //to Q in the Y[0] register.
//Y[5] = ((Y[6] & 0x7E0000) >> (9 + 8)) | ((Y[6] & 0x7E000000) >> (9 + 16 - 6)) | ((Y[6] & 0x7E00000000) >> (9 + 24 - 12)) | ((Y[6] & 0x7E0000000000) >> (9 + 32 - 18));
//Y[3] = ((Y[4] & 0x3C000000) >> (18 + 8)) | ((Y[4] & 0x3C00000000) >> (18 + 16 - 4));
//Y[15] is the output block where the data must be stored.
//Every block = 256 / 8 = 64 bytes =
Y[5] = Y[5] * 64;
//Y[3] is the inner 2x8 block, holding 8 bits, a number from 0 - 255. The upper 3 bits denote the dword to store the data in. The lower 5 bits are a shift mask denoting which bit to flip.
//
Y[4] = (1i64 << (Y[3] & 31i64)); //or mask.
Y[3] = (Y[3] >> 5) * 4; //dword offset //8*4 = 32 bits
Y[5] = Y[5] | Y[3];
//** remove this line!
//Y[5] = Y[5] & ((MemOutputSize)-1); //mask at 512MB, because we don't have more memory.
//** remove the above line !
Y[5] = Y[5] + (unsigned __int64)doutput;
/*debug*/atomicOr((unsigned int *)Y[5], (unsigned int)Y[4]);
}
void printDevProp(cudaDeviceProp devProp)
{
printf("%s\n", devProp.name);
printf("Major revision number: %d\n", devProp.major);
printf("Minor revision number: %d\n", devProp.minor);
printf("Total global memory: %zu", devProp.totalGlobalMem);
printf(" bytes\n");
printf("Number of multiprocessors: %d\n", devProp.multiProcessorCount);
printf("Total amount of shared memory per block: %zu\n", devProp.sharedMemPerBlock);
printf("Total registers per block: %d\n", devProp.regsPerBlock);
printf("Warp size: %d\n", devProp.warpSize);
printf("Maximum memory pitch: %zu\n", devProp.memPitch);
printf("Total amount of constant memory: %zu\n", devProp.totalConstMem);
return;
}
unsigned long long getTotalSystemMemory()
{
MEMORYSTATUSEX status;
status.dwLength = sizeof(status);
GlobalMemoryStatusEx(&status);
return status.ullTotalPhys;
}
#define filename "lookuptable5to3_doublecheckA.bin"
// Start the main CUDA Sample here
int main(int argc, char **argv)
{
printf("CUDA Lookup table 5x5->3x3 dual GPU version\n");
int deviceCount;
cudaGetDeviceCount(&deviceCount);
printf("Device count is %i, ", deviceCount);
printf("Available RAM = %lliGiB\n", (getTotalSystemMemory() >> 30i64));
if (deviceCount > 2) { deviceCount = 2; }
// This will pick the best possible CUDA capable device
int dev1 = findCudaDevice(argc, (const char **)argv);
cudaDeviceProp dp;
checkCudaErrors(cudaGetDeviceProperties(&dp, dev1));
printDevProp(dp);
//float *dinput = NULL;
unsigned long long int* dGPUoutput[2];
for (int dev = 0; dev < deviceCount; dev++) {
dGPUoutput[dev] = NULL;
}
//clock_t *dtimer = NULL;
//clock_t timer[NUM_BLOCKS * 2];
//float input[NUM_THREADS * 2];
//for (int i = 0; i < NUM_THREADS * 2; i++)
//{
// input[i] = (float)i;
//}
//unsigned __int64 a = 0xFFFFFFFFFFFFFFFF;
//unsigned __int64 b;
//SevenToFive(&a, &b);
//checkCudaErrors(cudaMalloc((void **)&dinput, sizeof(float) * NUM_THREADS * 2));
//checkCudaErrors(cudaMemcpy(dinput, input, sizeof(float) * NUM_THREADS * 2, cudaMemcpyHostToDevice));
//timedReduction<<<NUM_BLOCKS, NUM_THREADS, sizeof(float) * 2 *NUM_THREADS>>>(dinput, doutput, dtimer);
char* dest[2];
char* outputdest;
outputdest = (char*)malloc(sizeof(char) * MemOutputSize);
if (outputdest == NULL) { printf("Out of memory"); getch(); exit(EXIT_FAILURE); }
//test write
//printf("test write of data\n");
//std::ofstream outputFile;
//outputFile.open(filename, std::ofstream::out | std::ofstream::trunc | std::ofstream::binary);
///*debug*/outputFile.write(dest, MemOutputSize);
//outputFile.close();
printf("Start computing\n");
//DebugSevenToFive((unsigned long long int)-1, NULL); //just a place to check if needed.
//getch();
for (int dev = 0; dev < deviceCount; dev++) {
cudaSetDevice(dev);
/*debug*/checkCudaErrors(cudaMalloc((void **)&dGPUoutput[dev], sizeof(char) * MemOutputSize));
/*debug*/checkCudaErrors(cudaMemset(dGPUoutput[dev], 0, sizeof(char) * MemOutputSize));
dest[dev] = (char*)malloc(sizeof(char) * MemOutputSize);
if (dest[dev] == NULL) { printf("Out of memory"); getch(); exit(EXIT_FAILURE); }
}
auto t_start = std::chrono::high_resolution_clock::now();
//getch();
//we need to repeat this 2^31 times. 31=5+6+20
try {
for (int k = 0; k < 64; k++) { //2^6
printf("%i of 64 ", k + 1);
for (int j = 0; j < 32; j++) { //2^5
printf(".");
for (int q = 0; q < 1024; q++) {
printf("%4i\b\b\b\b", 1023 - q);
for (int i = 0; i < (/*1024 **/ 1024); i++) { //2^20
//__global__ static void SevenToFiveCount(const unsigned __int64 input, void* doutput)
///*debug*/SevenToFiveCount << <NUM_BLOCKS, NUM_THREADS >> > (i + (q * 1024) + (j * 1024 * 1024) + (k * 1024 * 1024 * 32), doutput); //256K, need to run this code 2GB times.
int dev = i % deviceCount;
cudaSetDevice(dev);
SevenToFive<<<NUM_BLOCKS, NUM_THREADS>>> (i + (q * 1024) + (j * 1024 * 1024) + (k * 1024 * 1024 * 32), dGPUoutput[dev]); //256K, need to run this code 2GB times.
}
}
}
printf("-");
for (int dev = 0; dev < deviceCount; dev++) {
cudaSetDevice(dev);
checkCudaErrors(cudaDeviceSynchronize());
/*debug*/checkCudaErrors(cudaMemcpy(dest[dev], dGPUoutput[dev], sizeof(char) * MemOutputSize, cudaMemcpyDeviceToHost));
}
for (__int64 i = 0; i < MemOutputSize * sizeof(char); i++) {
outputdest[i] = dest[0][i];
for (int dev = 1; dev < deviceCount; dev++) {
outputdest[i] |= dest[dev][i];
}
}
std::ofstream outputFile;
outputFile.open(filename, std::ofstream::out | std::ofstream::trunc | std::ofstream::binary);
outputFile.write(outputdest, MemOutputSize);
outputFile.close();
printf("W \n");
}
cudaDeviceSynchronize();
printf("\nDone computing\n");
auto t_end = std::chrono::high_resolution_clock::now();
std::cout << "Millisecs used " << std::chrono::duration<double, std::milli>(t_end - t_start).count() << "ms\n";
int check = 0;
for (int i = 0; i < (1024 * 1024 * 1024 / 4); i++) {
check |= outputdest[i];
}
if (check == 0) { printf("Error: nothing happened"); getch(); for (;;) {} }
else { printf("all ok"); getch(); }
//checkCudaErrors(cudaFree(dinput));
for (int dev = 0; dev < deviceCount; dev++) {
cudaFree(dGPUoutput[dev]);
}
free(outputdest);
outputdest = NULL;
for (int dev = 0; dev < deviceCount; dev++) {
free(dest[dev]);
dest[dev] = NULL;
}
getch();
return EXIT_SUCCESS;
}
catch (const std::exception& e) //catch all exceptions
{
printf("Oops, an error happened. Here are the details:\n");
std::cout << e.what() << std::endl;
printf("\nPress any key\n");
getch();
}
return EXIT_FAILURE;
}
(this doesn't fit into a comment, sorry)
Prepare a solver, feed it all GPUs instantanous or average performances for last several seconds as variables. Minimize total time per N kernels or maximize 1 kernel / average time value.
If kernels are identical, there should be only 1 minima point such as 90 kernels to gtx 1080 and 10 kernels to gtx1050, per second. If there are many different kernels, then they could become other variables for the solver.
Solver includes some math you can do maybe. I'll add only a solverless version:
give them their own queues of kernels
have them get kernels one at a time (finishes, gets a new )
check size of those queues.
Empty queue = 1.0 performance.
Full queue = 0.0 performance
continue optimizing by "produce"ing a kernel to 1 queue at a time(maybe a fixed time or quickly until one of queues is filled), by probability equal to performance, probably a normalized version which makes 0.1 0.1 equal to 0.5 0.5. Set max size of each queue to 50 to 100 for simple accuracy of performance.
In CUDA you can assign multiple "stream"s so that you can do this with multiple queues per GPU, asynchronously to each other, to increase GPU usage further. It's important to check sum of queues performance values per GPU stays in tune with its real performance.
Since you run millions of kernels, you can have queues with thousands of kernels to make a better performance measurement. Just 5 max size of a queue may not be very good to share a few kernels.
Maybe you can make queue max sizes adapting to their attached GPU performances. Starting with just 5 max size, increasing to thousands on fastest GPUs so that performance can be measured like 0.5444533 instead of just 0.5.
When last kernel is enqueued, all remaining number of elements in queues should be equal to or near their own GPU performances and expected to finish at the same time.
(untested)

Persistent Byte Output Counter C++ [closed]

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I am working on a project in which I need to keep track of how many bytes the software dishes out.
The software will be turned on and off occasionally, so I must implement a way to store the number of bytes outputted in a way so that an Administrator or another User cannot simply open the file and change the number of bytes outputted.
What is the best way to implement this?
Also, I am not able to use any libraries (ex: boost).
Here is an example of using https://en.wikipedia.org/wiki/Tiny_Encryption_Algorithm
#include <iostream>
#include <fstream>
void encrypt(uint32_t* v, uint32_t* k) {
uint32_t v0 = v[0], v1 = v[1], sum = 0, i; /* set up */
uint32_t delta = 0x9e3779b9; /* a key schedule constant */
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; /* cache key */
for (i = 0; i < 32; i++) { /* basic cycle start */
sum += delta;
v0 += ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
v1 += ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
} /* end cycle */
v[0] = v0; v[1] = v1;
}
void decrypt(uint32_t* v, uint32_t* k) {
uint32_t v0 = v[0], v1 = v[1], sum = 0xC6EF3720, i; /* set up */
uint32_t delta = 0x9e3779b9; /* a key schedule constant */
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3]; /* cache key */
for (i = 0; i<32; i++) { /* basic cycle start */
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
sum -= delta;
} /* end cycle */
v[0] = v0; v[1] = v1;
}
int main()
{
uint32_t k[4] = { 123,456,789,10 }; // key
uint32_t v[2] = { 1000000, 1000000 }; // data
// save into file
std::ofstream ofs("save.dat", std::ios::binary);
encrypt(v, k);
ofs << v[0] << " " << v[1] << std::endl;
// read from file
std::ifstream ifs("save.dat", std::ios::binary);
uint32_t v2[2];
if (ifs >> v2[0] >> v2[1])
{
std::cout << "Filedata: " << v2[0] << " " << v2[1] << std::endl;
decrypt(v2, k);
if (v2[0] == v2[1])
std::cout << "Decrypted: " << v2[0] << std::endl;
else
std::cout << "Data was tampered with!" << std::endl;
}
}
http://coliru.stacked-crooked.com/view?id=d725bf798ff8ca12
Works pretty good and doesn't need any library. This is low level protection but should be hard enough to discourage your users.

Convert 0x1234 to 0x11223344

How do I expand the hexadecimal number 0x1234 to 0x11223344 in a high-performance way?
unsigned int c = 0x1234, b;
b = (c & 0xff) << 4 | c & 0xf | (c & 0xff0) << 8
| (c & 0xff00) << 12 | (c & 0xf000) << 16;
printf("%p -> %p\n", c, b);
Output:
0x1234 -> 0x11223344
I need this for color conversion. Users provide their data in the form 0xARGB, and I need to convert it to 0xAARRGGBB. And yes, there could be millions, because each could be a pixel. 1000x1000 pixels equals to one million.
The actual case is even more complicated, because a single 32-bit value contains both foreground and background colors. So 0xARGBargb become: [ 0xAARRGGBB, 0xaarrggbb ]
Oh yes, one more thing, in a real application I also negate alpha, because in OpenGL 0xFF is non-transparent and 0x00 is most transparent, which is inconvenient in most cases, because usually you just need an RGB part and transparency is assumed to be non-present.
This can be done using SSE2 as follows:
void ExpandSSE2(unsigned __int64 in, unsigned __int64 &outLo, unsigned __int64 &outHi) {
__m128i const mask = _mm_set1_epi16((short)0xF00F);
__m128i const mul0 = _mm_set1_epi16(0x0011);
__m128i const mul1 = _mm_set1_epi16(0x1000);
__m128i v;
v = _mm_cvtsi64_si128(in); // Move the 64-bit value to a 128-bit register
v = _mm_unpacklo_epi8(v, v); // 0x12 -> 0x1212
v = _mm_and_si128(v, mask); // 0x1212 -> 0x1002
v = _mm_mullo_epi16(v, mul0); // 0x1002 -> 0x1022
v = _mm_mulhi_epu16(v, mul1); // 0x1022 -> 0x0102
v = _mm_mullo_epi16(v, mul0); // 0x0102 -> 0x1122
outLo = _mm_extract_epi64(v, 0);
outHi = _mm_extract_epi64(v, 1);
}
Of course you’d want to put the guts of the function in an inner loop and pull out the constants. You will also want to skip the x64 registers and load values directly into 128-bit SSE registers. For an example of how to do this, refer to the SSE2 implementation in the performance test below.
At its core, there are five instructions, which perform the operation on four color values at a time. So, that is only about 1.25 instructions per color value. It should also be noted that SSE2 is available anywhere x64 is available.
Performance tests for an assortment of the solutions here
A few people have mentioned that the only way to know what's faster is to run the code, and this is unarguably true. So I've compiled a few of the solutions into a performance test so we can compare apples to apples. I chose solutions which I felt were significantly different from the others enough to require testing. All the solutions read from memory, operate on the data, and write back to memory. In practice some of the SSE solutions will require additional care around the alignment and handling cases when there aren't another full 16 bytes to process in the input data. The code I tested is x64 compiled under release using Visual Studio 2013 running on a 4+ GHz Core i7.
Here are my results:
ExpandOrig: 56.234 seconds // From asker's original question
ExpandSmallLUT: 30.209 seconds // From Dmitry's answer
ExpandLookupSmallOneLUT: 33.689 seconds // from Dmitry's answer
ExpandLookupLarge: 51.312 seconds // A straightforward lookup table
ExpandAShelly: 43.829 seconds // From AShelly's answer
ExpandAShellyMulOp: 43.580 seconds // AShelly's answer with an optimization
ExpandSSE4: 17.854 seconds // My original SSE4 answer
ExpandSSE4Unroll: 17.405 seconds // My original SSE4 answer with loop unrolling
ExpandSSE2: 17.281 seconds // My current SSE2 answer
ExpandSSE2Unroll: 17.152 seconds // My current SSE2 answer with loop unrolling
In the test results above you'll see I included the asker's code, three lookup table implementations including the small lookup table implementation proposed in Dmitry's answer. AShelly's solution is included too, as well as a version with an optimization I made (an operation can be eliminated). I included my original SSE4 implementation, as well as a superior SSE2 version I made later (now reflected as the answer), as well as unrolled versions of both since they were the fastest here, and I wanted to see how much unrolling sped them up. I also included an SSE4 implementation of AShelly's answer.
So far I have to declare myself the winner. But the source is below, so anyone can test it out on their platform, and include their own solution into the testing to see if they've made a solution that's even faster.
#define DATA_SIZE_IN ((unsigned)(1024 * 1024 * 128))
#define DATA_SIZE_OUT ((unsigned)(2 * DATA_SIZE_IN))
#define RERUN_COUNT 500
#include <cstdlib>
#include <ctime>
#include <iostream>
#include <utility>
#include <emmintrin.h> // SSE2
#include <tmmintrin.h> // SSSE3
#include <smmintrin.h> // SSE4
void ExpandOrig(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
u = (u & 0x00FF) << 4
| (u & 0x000F)
| (u & 0x0FF0) << 8
| (u & 0xFF00) << 12
| (u & 0xF000) << 16;
v = (v & 0x00FF) << 4
| (v & 0x000F)
| (v & 0x0FF0) << 8
| (v & 0xFF00) << 12
| (v & 0xF000) << 16;
// Store data
*(unsigned*)(out) = u;
*(unsigned*)(out + 4) = v;
in += 4;
out += 8;
} while (in != past);
}
unsigned LutLo[256],
LutHi[256];
void MakeLutLo(void) {
for (unsigned i = 0, x; i < 256; ++i) {
x = i;
x = ((x & 0xF0) << 4) | (x & 0x0F);
x |= (x << 4);
LutLo[i] = x;
}
}
void MakeLutHi(void) {
for (unsigned i = 0, x; i < 256; ++i) {
x = i;
x = ((x & 0xF0) << 20) | ((x & 0x0F) << 16);
x |= (x << 4);
LutHi[i] = x;
}
}
void ExpandLookupSmall(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
u = LutHi[u >> 8] | LutLo[u & 0xFF];
v = LutHi[v >> 8] | LutLo[v & 0xFF];
// Store data
*(unsigned*)(out) = u;
*(unsigned*)(out + 4) = v;
in += 4;
out += 8;
} while (in != past);
}
void ExpandLookupSmallOneLUT(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
u = ((LutLo[u >> 8] << 16) | LutLo[u & 0xFF]);
v = ((LutLo[v >> 8] << 16) | LutLo[v & 0xFF]);
// Store data
*(unsigned*)(out) = u;
*(unsigned*)(out + 4) = v;
in += 4;
out += 8;
} while (in != past);
}
unsigned LutLarge[256 * 256];
void MakeLutLarge(void) {
for (unsigned i = 0; i < (256 * 256); ++i)
LutLarge[i] = LutHi[i >> 8] | LutLo[i & 0xFF];
}
void ExpandLookupLarge(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
u = LutLarge[u];
v = LutLarge[v];
// Store data
*(unsigned*)(out) = u;
*(unsigned*)(out + 4) = v;
in += 4;
out += 8;
} while (in != past);
}
void ExpandAShelly(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v, w, x;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
w = (((u & 0xF0F) * 0x101) & 0xF000F) + (((u & 0xF0F0) * 0x1010) & 0xF000F00);
x = (((v & 0xF0F) * 0x101) & 0xF000F) + (((v & 0xF0F0) * 0x1010) & 0xF000F00);
w += w * 0x10;
x += x * 0x10;
// Store data
*(unsigned*)(out) = w;
*(unsigned*)(out + 4) = x;
in += 4;
out += 8;
} while (in != past);
}
void ExpandAShellyMulOp(unsigned char const *in, unsigned char const *past, unsigned char *out) {
unsigned u, v;
do {
// Read in data
u = *(unsigned const*)in;
v = u >> 16;
u &= 0x0000FFFF;
// Do computation
u = ((((u & 0xF0F) * 0x101) & 0xF000F) + (((u & 0xF0F0) * 0x1010) & 0xF000F00)) * 0x11;
v = ((((v & 0xF0F) * 0x101) & 0xF000F) + (((v & 0xF0F0) * 0x1010) & 0xF000F00)) * 0x11;
// Store data
*(unsigned*)(out) = u;
*(unsigned*)(out + 4) = v;
in += 4;
out += 8;
} while (in != past);
}
void ExpandSSE4(unsigned char const *in, unsigned char const *past, unsigned char *out) {
__m128i const mask0 = _mm_set1_epi16((short)0x8000),
mask1 = _mm_set1_epi8(0x0F),
mul = _mm_set1_epi16(0x0011);
__m128i u, v, w, x;
do {
// Read input into low 8 bytes of u and v
u = _mm_load_si128((__m128i const*)in);
v = _mm_unpackhi_epi8(u, u); // Expand each single byte to two bytes
u = _mm_unpacklo_epi8(u, u); // Do it again for v
w = _mm_srli_epi16(u, 4); // Copy the value into w and shift it right half a byte
x = _mm_srli_epi16(v, 4); // Do it again for v
u = _mm_blendv_epi8(u, w, mask0); // Select odd bytes from w, and even bytes from v, giving the the desired value in the upper nibble of each byte
v = _mm_blendv_epi8(v, x, mask0); // Do it again for v
u = _mm_and_si128(u, mask1); // Clear the all the upper nibbles
v = _mm_and_si128(v, mask1); // Do it again for v
u = _mm_mullo_epi16(u, mul); // Multiply each 16-bit value by 0x0011 to duplicate the lower nibble in the upper nibble of each byte
v = _mm_mullo_epi16(v, mul); // Do it again for v
// Write output
_mm_store_si128((__m128i*)(out ), u);
_mm_store_si128((__m128i*)(out + 16), v);
in += 16;
out += 32;
} while (in != past);
}
void ExpandSSE4Unroll(unsigned char const *in, unsigned char const *past, unsigned char *out) {
__m128i const mask0 = _mm_set1_epi16((short)0x8000),
mask1 = _mm_set1_epi8(0x0F),
mul = _mm_set1_epi16(0x0011);
__m128i u0, v0, w0, x0,
u1, v1, w1, x1,
u2, v2, w2, x2,
u3, v3, w3, x3;
do {
// Read input into low 8 bytes of u and v
u0 = _mm_load_si128((__m128i const*)(in ));
u1 = _mm_load_si128((__m128i const*)(in + 16));
u2 = _mm_load_si128((__m128i const*)(in + 32));
u3 = _mm_load_si128((__m128i const*)(in + 48));
v0 = _mm_unpackhi_epi8(u0, u0); // Expand each single byte to two bytes
u0 = _mm_unpacklo_epi8(u0, u0); // Do it again for v
v1 = _mm_unpackhi_epi8(u1, u1); // Do it again
u1 = _mm_unpacklo_epi8(u1, u1); // Again for u1
v2 = _mm_unpackhi_epi8(u2, u2); // Again for v1
u2 = _mm_unpacklo_epi8(u2, u2); // Again for u2
v3 = _mm_unpackhi_epi8(u3, u3); // Again for v2
u3 = _mm_unpacklo_epi8(u3, u3); // Again for u3
w0 = _mm_srli_epi16(u0, 4); // Copy the value into w and shift it right half a byte
x0 = _mm_srli_epi16(v0, 4); // Do it again for v
w1 = _mm_srli_epi16(u1, 4); // Again for u1
x1 = _mm_srli_epi16(v1, 4); // Again for v1
w2 = _mm_srli_epi16(u2, 4); // Again for u2
x2 = _mm_srli_epi16(v2, 4); // Again for v2
w3 = _mm_srli_epi16(u3, 4); // Again for u3
x3 = _mm_srli_epi16(v3, 4); // Again for v3
u0 = _mm_blendv_epi8(u0, w0, mask0); // Select even bytes from w, and odd bytes from v, giving the the desired value in the upper nibble of each byte
v0 = _mm_blendv_epi8(v0, x0, mask0); // Do it again for v
u1 = _mm_blendv_epi8(u1, w1, mask0); // Again for u1
v1 = _mm_blendv_epi8(v1, x1, mask0); // Again for v1
u2 = _mm_blendv_epi8(u2, w2, mask0); // Again for u2
v2 = _mm_blendv_epi8(v2, x2, mask0); // Again for v2
u3 = _mm_blendv_epi8(u3, w3, mask0); // Again for u3
v3 = _mm_blendv_epi8(v3, x3, mask0); // Again for v3
u0 = _mm_and_si128(u0, mask1); // Clear the all the upper nibbles
v0 = _mm_and_si128(v0, mask1); // Do it again for v
u1 = _mm_and_si128(u1, mask1); // Again for u1
v1 = _mm_and_si128(v1, mask1); // Again for v1
u2 = _mm_and_si128(u2, mask1); // Again for u2
v2 = _mm_and_si128(v2, mask1); // Again for v2
u3 = _mm_and_si128(u3, mask1); // Again for u3
v3 = _mm_and_si128(v3, mask1); // Again for v3
u0 = _mm_mullo_epi16(u0, mul); // Multiply each 16-bit value by 0x0011 to duplicate the lower nibble in the upper nibble of each byte
v0 = _mm_mullo_epi16(v0, mul); // Do it again for v
u1 = _mm_mullo_epi16(u1, mul); // Again for u1
v1 = _mm_mullo_epi16(v1, mul); // Again for v1
u2 = _mm_mullo_epi16(u2, mul); // Again for u2
v2 = _mm_mullo_epi16(v2, mul); // Again for v2
u3 = _mm_mullo_epi16(u3, mul); // Again for u3
v3 = _mm_mullo_epi16(v3, mul); // Again for v3
// Write output
_mm_store_si128((__m128i*)(out ), u0);
_mm_store_si128((__m128i*)(out + 16), v0);
_mm_store_si128((__m128i*)(out + 32), u1);
_mm_store_si128((__m128i*)(out + 48), v1);
_mm_store_si128((__m128i*)(out + 64), u2);
_mm_store_si128((__m128i*)(out + 80), v2);
_mm_store_si128((__m128i*)(out + 96), u3);
_mm_store_si128((__m128i*)(out + 112), v3);
in += 64;
out += 128;
} while (in != past);
}
void ExpandSSE2(unsigned char const *in, unsigned char const *past, unsigned char *out) {
__m128i const mask = _mm_set1_epi16((short)0xF00F),
mul0 = _mm_set1_epi16(0x0011),
mul1 = _mm_set1_epi16(0x1000);
__m128i u, v;
do {
// Read input into low 8 bytes of u and v
u = _mm_load_si128((__m128i const*)in);
v = _mm_unpackhi_epi8(u, u); // Expand each single byte to two bytes
u = _mm_unpacklo_epi8(u, u); // Do it again for v
u = _mm_and_si128(u, mask);
v = _mm_and_si128(v, mask);
u = _mm_mullo_epi16(u, mul0);
v = _mm_mullo_epi16(v, mul0);
u = _mm_mulhi_epu16(u, mul1); // This can also be done with a right shift of 4 bits, but this seems to mesure faster
v = _mm_mulhi_epu16(v, mul1);
u = _mm_mullo_epi16(u, mul0);
v = _mm_mullo_epi16(v, mul0);
// write output
_mm_store_si128((__m128i*)(out ), u);
_mm_store_si128((__m128i*)(out + 16), v);
in += 16;
out += 32;
} while (in != past);
}
void ExpandSSE2Unroll(unsigned char const *in, unsigned char const *past, unsigned char *out) {
__m128i const mask = _mm_set1_epi16((short)0xF00F),
mul0 = _mm_set1_epi16(0x0011),
mul1 = _mm_set1_epi16(0x1000);
__m128i u0, v0,
u1, v1;
do {
// Read input into low 8 bytes of u and v
u0 = _mm_load_si128((__m128i const*)(in ));
u1 = _mm_load_si128((__m128i const*)(in + 16));
v0 = _mm_unpackhi_epi8(u0, u0); // Expand each single byte to two bytes
u0 = _mm_unpacklo_epi8(u0, u0); // Do it again for v
v1 = _mm_unpackhi_epi8(u1, u1); // Do it again
u1 = _mm_unpacklo_epi8(u1, u1); // Again for u1
u0 = _mm_and_si128(u0, mask);
v0 = _mm_and_si128(v0, mask);
u1 = _mm_and_si128(u1, mask);
v1 = _mm_and_si128(v1, mask);
u0 = _mm_mullo_epi16(u0, mul0);
v0 = _mm_mullo_epi16(v0, mul0);
u1 = _mm_mullo_epi16(u1, mul0);
v1 = _mm_mullo_epi16(v1, mul0);
u0 = _mm_mulhi_epu16(u0, mul1);
v0 = _mm_mulhi_epu16(v0, mul1);
u1 = _mm_mulhi_epu16(u1, mul1);
v1 = _mm_mulhi_epu16(v1, mul1);
u0 = _mm_mullo_epi16(u0, mul0);
v0 = _mm_mullo_epi16(v0, mul0);
u1 = _mm_mullo_epi16(u1, mul0);
v1 = _mm_mullo_epi16(v1, mul0);
// write output
_mm_store_si128((__m128i*)(out ), u0);
_mm_store_si128((__m128i*)(out + 16), v0);
_mm_store_si128((__m128i*)(out + 32), u1);
_mm_store_si128((__m128i*)(out + 48), v1);
in += 32;
out += 64;
} while (in != past);
}
void ExpandAShellySSE4(unsigned char const *in, unsigned char const *past, unsigned char *out) {
__m128i const zero = _mm_setzero_si128(),
v0F0F = _mm_set1_epi32(0x0F0F),
vF0F0 = _mm_set1_epi32(0xF0F0),
v0101 = _mm_set1_epi32(0x0101),
v1010 = _mm_set1_epi32(0x1010),
v000F000F = _mm_set1_epi32(0x000F000F),
v0F000F00 = _mm_set1_epi32(0x0F000F00),
v0011 = _mm_set1_epi32(0x0011);
__m128i u, v, w, x;
do {
// Read in data
u = _mm_load_si128((__m128i const*)in);
v = _mm_unpackhi_epi16(u, zero);
u = _mm_unpacklo_epi16(u, zero);
// original source: ((((a & 0xF0F) * 0x101) & 0xF000F) + (((a & 0xF0F0) * 0x1010) & 0xF000F00)) * 0x11;
w = _mm_and_si128(u, v0F0F);
x = _mm_and_si128(v, v0F0F);
u = _mm_and_si128(u, vF0F0);
v = _mm_and_si128(v, vF0F0);
w = _mm_mullo_epi32(w, v0101); // _mm_mullo_epi32 is what makes this require SSE4 instead of SSE2
x = _mm_mullo_epi32(x, v0101);
u = _mm_mullo_epi32(u, v1010);
v = _mm_mullo_epi32(v, v1010);
w = _mm_and_si128(w, v000F000F);
x = _mm_and_si128(x, v000F000F);
u = _mm_and_si128(u, v0F000F00);
v = _mm_and_si128(v, v0F000F00);
u = _mm_add_epi32(u, w);
v = _mm_add_epi32(v, x);
u = _mm_mullo_epi32(u, v0011);
v = _mm_mullo_epi32(v, v0011);
// write output
_mm_store_si128((__m128i*)(out ), u);
_mm_store_si128((__m128i*)(out + 16), v);
in += 16;
out += 32;
} while (in != past);
}
int main() {
unsigned char *const indat = new unsigned char[DATA_SIZE_IN ],
*const outdat0 = new unsigned char[DATA_SIZE_OUT],
*const outdat1 = new unsigned char[DATA_SIZE_OUT],
* curout = outdat0,
* lastout = outdat1,
* place;
unsigned start,
stop;
place = indat + DATA_SIZE_IN - 1;
do {
*place = (unsigned char)rand();
} while (place-- != indat);
MakeLutLo();
MakeLutHi();
MakeLutLarge();
for (unsigned testcount = 0; testcount < 1000; ++testcount) {
// Solution posted by the asker
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandOrig(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandOrig:\t\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
// Dmitry's small lookup table solution
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandLookupSmall(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandSmallLUT:\t\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// Dmitry's small lookup table solution using only one lookup table
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandLookupSmallOneLUT(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandLookupSmallOneLUT:\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// Large lookup table solution
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandLookupLarge(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandLookupLarge:\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// AShelly's Interleave bits by Binary Magic Numbers solution
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandAShelly(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandAShelly:\t\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// AShelly's Interleave bits by Binary Magic Numbers solution optimizing out an addition
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandAShellyMulOp(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandAShellyMulOp:\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// My SSE4 solution
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandSSE4(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandSSE4:\t\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// My SSE4 solution unrolled
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandSSE4Unroll(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandSSE4Unroll:\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// My SSE2 solution
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandSSE2(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandSSE2:\t\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// My SSE2 solution unrolled
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandSSE2Unroll(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandSSE2Unroll:\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
// AShelly's Interleave bits by Binary Magic Numbers solution implemented using SSE2
start = clock();
for (unsigned rerun = 0; rerun < RERUN_COUNT; ++rerun)
ExpandAShellySSE4(indat, indat + DATA_SIZE_IN, curout);
stop = clock();
std::cout << "ExpandAShellySSE4:\t\t" << (((stop - start) / 1000) / 60) << ':' << (((stop - start) / 1000) % 60) << ":." << ((stop - start) % 1000) << std::endl;
std::swap(curout, lastout);
if (memcmp(outdat0, outdat1, DATA_SIZE_OUT))
std::cout << "INCORRECT OUTPUT" << std::endl;
}
delete[] indat;
delete[] outdat0;
delete[] outdat1;
return 0;
}
NOTE:
I had an SSE4 implementation here initially. I found a way to implement this using SSE2, which is better because it will run on more platforms. The SSE2 implementation is also faster. So, the solution presented at the top is now the SSE2 implementation and not the SSE4 one. The SSE4 implementation can still be seen in the performance tests or in the edit history.
I'm not sure what the most efficient way would be, but this is a little shorter:
#include <stdio.h>
int main()
{
unsigned x = 0x1234;
x = (x << 8) | x;
x = ((x & 0x00f000f0) << 4) | (x & 0x000f000f);
x = (x << 4) | x;
printf("0x1234 -> 0x%08x\n",x);
return 0;
}
If you need to do this repeatedly and very quickly, as suggested in your edit, you could consider generating a lookup table and using that instead. The following function dynamically allocates and initializes such a table:
unsigned *makeLookupTable(void)
{
unsigned *tbl = malloc(sizeof(unsigned) * 65536);
if (!tbl) return NULL;
int i;
for (i = 0; i < 65536; i++) {
unsigned x = i;
x |= (x << 8);
x = ((x & 0x00f000f0) << 4) | (x & 0x000f000f);
x |= (x << 4);
/* Uncomment next line to invert the high byte as mentioned in the edit. */
/* x = x ^ 0xff000000; */
tbl[i] = x;
}
return tbl;
}
After that each conversion is just something like:
result = lookuptable[input];
..or maybe:
result = lookuptable[input & 0xffff];
Or a smaller, more cache-friendly lookup table (or pair) could be used with one lookup each for the high and low bytes (as noted by #LưuVĩnhPhúc in the comments). In that case, table generation code might be:
unsigned *makeLookupTableLow(void)
{
unsigned *tbl = malloc(sizeof(unsigned) * 256);
if (!tbl) return NULL;
int i;
for (i = 0; i < 256; i++) {
unsigned x = i;
x = ((x & 0xf0) << 4) | (x & 0x0f);
x |= (x << 4);
tbl[i] = x;
}
return tbl;
}
...and an optional second table:
unsigned *makeLookupTableHigh(void)
{
unsigned *tbl = malloc(sizeof(unsigned) * 256);
if (!tbl) return NULL;
int i;
for (i = 0; i < 256; i++) {
unsigned x = i;
x = ((x & 0xf0) << 20) | ((x & 0x0f) << 16);
x |= (x << 4);
/* uncomment next line to invert high byte */
/* x = x ^ 0xff000000; */
tbl[i] = x;
}
return tbl;
}
...and to convert a value with two tables:
result = hightable[input >> 8] | lowtable[input & 0xff];
...or with one (just the low table above):
result = (lowtable[input >> 8] << 16) | lowtable[input & 0xff];
result ^= 0xff000000; /* to invert high byte */
If the upper part of the value (alpha?) doesn't change much, even the single large table might perform well since consecutive lookups would be closer together in the table.
I took the performance test code #Apriori posted, made some adjustments, and added tests for the other responses that he hadn't included originally... then compiled three versions of it with different settings. One is 64-bit code with SSE4.1 enabled, where the compiler can make use of SSE for optimizations... and then two 32-bit versions, one with SSE and one without. Although all three were run on the same fairly recent processor, the results show how the optimal solution can change depending on the processor features:
64b SSE4.1 32b SSE4.1 32b no SSE
-------------------------- ---------- ---------- ----------
ExpandOrig time: 3.502 s 3.501 s 6.260 s
ExpandLookupSmall time: 3.530 s 3.997 s 3.996 s
ExpandLookupLarge time: 3.434 s 3.419 s 3.427 s
ExpandIsalamon time: 3.654 s 3.673 s 8.870 s
ExpandIsalamonOpt time: 3.784 s 3.720 s 8.719 s
ExpandChronoKitsune time: 3.658 s 3.463 s 6.546 s
ExpandEvgenyKluev time: 6.790 s 7.697 s 13.383 s
ExpandIammilind time: 3.485 s 3.498 s 6.436 s
ExpandDmitri time: 3.457 s 3.477 s 5.461 s
ExpandNitish712 time: 3.574 s 3.800 s 6.789 s
ExpandAdamLiss time: 3.673 s 5.680 s 6.969 s
ExpandAShelly time: 3.524 s 4.295 s 5.867 s
ExpandAShellyMulOp time: 3.527 s 4.295 s 5.852 s
ExpandSSE4 time: 3.428 s
ExpandSSE4Unroll time: 3.333 s
ExpandSSE2 time: 3.392 s
ExpandSSE2Unroll time: 3.318 s
ExpandAShellySSE4 time: 3.392 s
The executables were compiled on 64-bit Linux with gcc 4.8.1, using -m64 -O3 -march=core2 -msse4.1, -m32 -O3 -march=core2 -msse4.1 and -m32 -O3 -march=core2 -mno-sse respectively. #Apriori's SSE tests were omitted for the 32-bit builds (crashed on 32-bit with SSE enabled, and obviously won't work with SSE disabled).
Among the adjustments made was to use actual image data instead of random values (photos of objects with transparent backgrounds), which greatly improved the performance of the large lookup table but made little difference for the others.
Essentially, the lookup tables win by a landslide when SSE is unnavailable (or unused)... and the manually coded SSE solutions win otherwise. However, it's also noteworthy that when the compiler could use SSE for optimizations, most of the bit manipulation solutions were almost as fast as the manually coded SSE -- still slower, but only marginally.
Here's another attempt, using eight operations:
b = (((c & 0x0F0F) * 0x0101) & 0x00F000F) +
(((c & 0xF0F0) * 0x1010) & 0xF000F00);
b += b * 0x10;
printf("%x\n",b); //Shows '0x11223344'
*Note, this post originally contained quite different code, based on Interleave bits by Binary Magic Numbers from Sean Anderson's bithacks page. But that wasn't quite what the OP was asking. so it has ben removed. The majority of the comments below refer to that missing version.
I wanted to add this link into the answer pool because I think it is extremely important when talking about optimization, to remember the hardware we are running on, as well as the technologies compiling our code for said platform.
Blog post Playing with the CPU pipeline is about looking into optimizing a set of code for the CPU pipelining. It actually shows an example of where he tries to simplify the math down to the fewest actual mathematical operations, yet it was FAR from the most optimal solution in terms of time. I have seen a couple of answers here speaking to that effect, and they may be correct, they may not. The only way to know is to actually measure the time from start to finish of your particular snippet of code, in comparison to others. Read this blog; it is EXTREMELY interesting.
I think I should mention that I am in this particular case not going to put ANY code up here unless I have truly tried multiple attempts, and actually gotten on that is particularly faster through multiple tries.
I think that the lookup table approach suggested by Dimitri is a good choice, but I suggest to go one step further and generate the table in compile time; doing the work at compile time will obviously lessen the execution time.
First, we create a compile-time value, using any of the suggested methods:
constexpr unsigned int transform1(unsigned int x)
{
return ((x << 8) | x);
}
constexpr unsigned int transform2(unsigned int x)
{
return (((x & 0x00f000f0) << 4) | (x & 0x000f000f));
}
constexpr unsigned int transform3(unsigned int x)
{
return ((x << 4) | x);
}
constexpr unsigned int transform(unsigned int x)
{
return transform3(transform2(transform1(x)));
}
// Dimitri version, using constexprs
template <unsigned int argb> struct aarrggbb_dimitri
{
static const unsigned int value = transform(argb);
};
// Adam Liss version
template <unsigned int argb> struct aarrggbb_adamLiss
{
static const unsigned int value =
(argb & 0xf000) * 0x11000 +
(argb & 0x0f00) * 0x01100 +
(argb & 0x00f0) * 0x00110 +
(argb & 0x000f) * 0x00011;
};
And then, we create the compile-time lookup table with whatever method we have available, I'll wish to use the C++14 integer sequence but I don't know which compiler will the OP be using. So another possible approach would be to use a pretty ugly macro:
#define EXPAND16(x) aarrggbb<x + 0>::value, \
aarrggbb<x + 1>::value, \
aarrggbb<x + 2>::value, \
aarrggbb<x + 3>::value, \
aarrggbb<x + 4>::value, \
aarrggbb<x + 5>::value, \
aarrggbb<x + 6>::value, \
... and so on
#define EXPAND EXPAND16(0), \
EXPAND16(0x10), \
EXPAND16(0x20), \
EXPAND16(0x30), \
EXPAND16(0x40), \
... and so on
... and so on
See demo here.
PS: The Adam Liss approach could be used without C++11.
If multiplication is cheap and 64-bit arithmetics is available, you could use this code:
uint64_t x = 0x1234;
x *= 0x0001000100010001ull;
x &= 0xF0000F0000F0000Full;
x *= 0x0000001001001001ull;
x &= 0xF0F0F0F000000000ull;
x = (x >> 36) * 0x11;
std::cout << std::hex << x << '\n';
In fact, it uses the same idea as the original attempt by AShelly.
This works and may be easier to understand, but bit manipulations are so cheap that I wouldn't worry much about efficiency.
#include <stdio.h>
#include <stdlib.h>
void main() {
unsigned int c = 0x1234, b;
b = (c & 0xf000) * 0x11000 + (c & 0x0f00) * 0x01100 +
(c & 0x00f0) * 0x00110 + (c & 0x000f) * 0x00011;
printf("%x -> %x\n", c, b);
}
Assuming that, you want to always convert 0xWXYZ to 0xWWXXYYZZ, I believe that below solution would be little faster than the one you suggested:
unsigned int c = 0x1234;
unsigned int b = (c & 0xf) | ((c & 0xf0) << 4) |
((c & 0xf00) << 8) | ((c & 0xf000) << 12);
b |= (b << 4);
Notice that, one &(and) operation is saved from your solution. :-)
Demo.
Another way is:
DWORD OrVal(DWORD & nible_pos, DWORD input_val, DWORD temp_val, int shift)
{
if (nible_pos==0)
nible_pos = 0x0000000F;
else
nible_pos = nible_pos << 4;
DWORD nible = input_val & nible_pos;
temp_val |= (nible << shift);
temp_val |= (nible << (shift + 4));
return temp_val;
}
DWORD Converter2(DWORD input_val)
{
DWORD nible_pos = 0x00000000;
DWORD temp_val = 0x00000000;
temp_val = OrVal(nible_pos, input_val, temp_val, 0);
temp_val = OrVal(nible_pos, input_val, temp_val, 4);
temp_val = OrVal(nible_pos, input_val, temp_val, 8);
temp_val = OrVal(nible_pos, input_val, temp_val, 12);
return temp_val;
}
DWORD val2 = Converter2(0x1234);
An optimized version (3 times faster):
DWORD Converter3(DWORD input_val)
{
DWORD nible_pos = 0;
DWORD temp_val = 0;
int shift = 0;
DWORD bit_nible[4] = { 0x000F, 0x000F0, 0x0F00, 0xF000 };
for ( ; shift < 16; shift+=4 )
{
if (nible_pos==0)
nible_pos = 0x0000000F;
else
nible_pos = nible_pos << 4;
DWORD nible = input_val & nible_pos;
temp_val |= (nible << shift);
temp_val |= (nible << (shift + 4));
}
return temp_val;
}
Perhaps this could be more simpler & efficient.
unsigned int g = 0x1234;
unsigned int ans = 0;
ans = ( ( g & 0xf000 ) << 16) + ( (g & 0xf00 ) << 12)
+ ( ( g&0xf0 ) << 8) + ( ( g&0xf ) << 4);
ans = ( ans | ans>>4 );
printf("%p -> %p\n", g, ans);
unsigned long transform(unsigned long n)
{
/* n: 00AR
* 00GB
*/
n = ((n & 0xff00) << 8) | (n & 0x00ff);
/* n: 0AR0
* 0GB0
*/
n <<= 4;
/* n: AAR0
* GGB0
*/
n |= (n & 0x0f000f00L) << 4;
/* n: AARR
* GGBB
*/
n |= (n & 0x00f000f0L) >> 4;
return n;
}
The alpha and red components are shifted into the higher 2 bytes where they belong, and the result is then shifted left by 4 bits, resulting in every component being exactly where it needs to be.
With a form of 0AR0 0GB0, a bit mask and left-shift combination is OR'ed with the current value. This copies the A and G components to the position just left of them. The same thing is done for the R and B components, except in the opposite direction.
If you are going to do this for OpenGL, I suggest you to use a glTexImageXD function with type parameter set to GL_UNSIGNED_SHORT_4_4_4_4. Your OpenGL driver should do the rest. And about the transparency inversion you can always manipulate blending via the glBlendFunc and glBlendEquation functions.
While others operate on hard-core optimization...
Take this as your best bet:
std::string toAARRGGBB(const std::string &argb)
{
std::string ret("0x");
int start = 2; //"0x####";
// ^^ skipped
for (int i = start;i < argb.length(); ++i)
{
ret += argb[i];
ret += argb[i];
}
return ret;
}
int main()
{
std::string argb = toAARRGGBB("0xACED"); //!!!
}
Haha