Difference between bitshifting mask vs unsigned int - c++

For a project, I had to find the individual 8-bits of a unsigned int. I first tried bit-shifting the mask to find the numbers, but that didn't work, so I tried bit-shifting the value and it worked.
What's the difference between these two? Why didn't the first one work?
ExampleFunk(unsigned int value){
for (int i = 0; i < 4; i++) {
ExampleSubFunk(value & (0x00FF << (i * 8)));
}
}
ExampleFunk(unsigned int value){
for (int i = 0; i < 4; i++) {
ExampleSubFunk((value >> (i * 8)) & 0x00FF);
}
}

Take the value 0xAABBCCDD as an example.
The expression value & (0xFF << (i * 8)) assumes the values:
0xAABBCCDD & 0x000000FF = 0x000000DD
0xAABBCCDD & 0x0000FF00 = 0x0000CC00
0xAABBCCDD & 0x00FF0000 = 0x00BB0000
0xAABBCCDD & 0xFF000000 = 0xAA000000
While the expression (value >> (i * 8)) & 0xFF assumes the values:
0xAABBCCDD & 0x000000FF = 0x000000DD
0x00AABBCC & 0x000000FF = 0x000000CC
0x0000AABB & 0x000000FF = 0x000000BB
0x000000AA & 0x000000FF = 0x000000AA
As you can see, the results are quite different after i = 0, because the first expression is only "selecting" 8 bits from value, while the second expression is shifting them down to the least significant byte first.
Note that in the first case, the expression (0xFF << (i * 8)) is shifting an int literal (0xFF) left. You should cast the literal to unsigned int to avoid signed integer overflow, which is undefined behavior:
value & ((unsigned int)0xFF << (i * 8))

In this code:
ExampleFunk(unsigned int value){
for (int i = 0; i < 4; i++) {
ExampleSubFunk(value & (0x00FF << (i * 8)));
}
}
You are shifting the bits of 0x00FF itself, producing new masks of 0x00FF, 0xFF00, 0xFF0000, and 0xFF000000, and then you are masking value with each of those masks. The result contains only the 8 bits of value that you are interested in, but those 8 bits are not moving position at all.
In this code:
ExampleFunk(unsigned int value){
for (int i = 0; i < 4; i++) {
ExampleSubFunk((value >> (i * 8)) & 0x00FF);
}
}
You are shifting the bits of value, thus moving those 8 bits that you want, and then you are masking the result with 0x00FF to extract those 8 bits.

Related

c++: how to put relevant bits from uint32 into uint8?

I have a uint32 that I've flagged some bits on:
uint32 i = 0;
i |= (1 << 0);
i |= (1 << 5);
i |= (1 << 13);
i |= (1 << 19);
...
I want to convert it to a uint8 (by getting the state of its first 8 bits and disregarding the rest). Obviously I could do this:
uint8 j = 0;
for (int q = 0; q < 8; q++)
{
if (i & (1 << q))
{
j |= (1 << q);
}
}
But is there a fancy bitwise operation I can use to transfer the bits over in one fell swoop, without a loop?
You can achieve the same result by simply assigning the uint32 value to uint8.
int main()
{
unsigned int i = 0x00000888;
unsigned char j = i;
cout<<hex<<i<<endl;
cout<<hex<<+j<<endl;
return 0;
}
output:
888
88
Why not just mask those last 8 bits instead of running a loop over to see if individual bits are set?
const unsigned char bitMask = 0xFF;
j = (i & bitMask);
Note that C++ 14 though allows you to define binary literals right away
const unsigned char bitMask = 0b1111'1111;
The above is all you need. Just in case, if you need to get the subsequent byte positions, use the same mask 0xFF and make sure to right shift back the result to get the desired byte value.

How to split an unsigned long int (32 bit) into 8 nibbles?

I am sorry if my question is confusing but here is the example of what I want to do,
lets say I have an unsigned long int = 1265985549
in binary I can write this as 01001011011101010110100000001101
now I want to split this binary 32 bit number into 4 bits like this and work separately on those 4 bits
0100 1011 0111 0101 0110 1000 0000 1101
any help would be appreciated.
You can get a 4-bit nibble at position k using bit operations, like this:
uint32_t nibble(uint32_t val, int k) {
return (val >> (4*k)) & 0x0F;
}
Now you can get the individual nibbles in a loop, like this:
uint32_t val = 1265985549;
for (int k = 0; k != 8 ; k++) {
uint32_t n = nibble(val, k);
cout << n << endl;
}
Demo on ideone.
short nibble0 = (i >> 0) & 15;
short nibble1 = (i >> 4) & 15;
short nibble2 = (i >> 8) & 15;
short nibble3 = (i >> 12) & 15;
etc
Based on the comment explaining the actual use for this, here's an other way to count how many nibbles have an odd parity: (not tested)
; compute parities of nibbles
x ^= x >> 2;
x ^= x >> 1;
x &= 0x11111111;
; add the parities
x = (x + (x >> 4)) & 0x0F0F0F0F;
int count = x * 0x01010101 >> 24;
The first part is just a regular "xor all the bits" type of parity calculation (where "all bits" refers to all the bits in a nibble, not in the entire integer), the second part is based on this bitcount algorithm, skipping some steps that are unnecessary because certain bits are always zero and so don't have to be added.

Extract n most significant non-zero bits from int in C++ without loops

I want to extract the n most significant bits from an integer in C++ and convert those n bits to an integer.
For example
int a=1200;
// its binary representation within 32 bit word-size is
// 00000000000000000000010010110000
Now I want to extract the 4 most significant digits from that representation, i.e. 1111
00000000000000000000010010110000
^^^^
and convert them again to an integer (1001 in decimal = 9).
How is possible with a simple c++ function without loops?
Some processors have an instruction to count the leading binary zeros of an integer, and some compilers have instrinsics to allow you to use that instruction. For example, using GCC:
uint32_t significant_bits(uint32_t value, unsigned bits) {
unsigned leading_zeros = __builtin_clz(value);
unsigned highest_bit = 32 - leading_zeros;
unsigned lowest_bit = highest_bit - bits;
return value >> lowest_bit;
}
For simplicity, I left out checks that the requested number of bits are available. For Microsoft's compiler, the intrinsic is called __lzcnt.
If your compiler doesn't provide that intrinsic, and you processor doesn't have a suitable instruction, then one way to count the zeros quickly is with a binary search:
unsigned leading_zeros(int32_t value) {
unsigned count = 0;
if ((value & 0xffff0000u) == 0) {
count += 16;
value <<= 16;
}
if ((value & 0xff000000u) == 0) {
count += 8;
value <<= 8;
}
if ((value & 0xf0000000u) == 0) {
count += 4;
value <<= 4;
}
if ((value & 0xc0000000u) == 0) {
count += 2;
value <<= 2;
}
if ((value & 0x80000000u) == 0) {
count += 1;
}
return count;
}
It's not fast, but (int)(log(x)/log(2) + .5) + 1 will tell you the position of the most significant non-zero bit. Finishing the algorithm from there is fairly straight-forward.
This seems to work (done in C# with UInt32 then ported so apologies to Bjarne):
unsigned int input = 1200;
unsigned int most_significant_bits_to_get = 4;
// shift + or the msb over all the lower bits
unsigned int m1 = input | input >> 8 | input >> 16 | input >> 24;
unsigned int m2 = m1 | m1 >> 2 | m1 >> 4 | m1 >> 6;
unsigned int m3 = m2 | m2 >> 1;
unsigned int nbitsmask = m3 ^ m3 >> most_significant_bits_to_get;
unsigned int v = nbitsmask;
unsigned int c = 32; // c will be the number of zero bits on the right
v &= -((int)v);
if (v>0) c--;
if ((v & 0x0000FFFF) >0) c -= 16;
if ((v & 0x00FF00FF) >0) c -= 8;
if ((v & 0x0F0F0F0F) >0 ) c -= 4;
if ((v & 0x33333333) >0) c -= 2;
if ((v & 0x55555555) >0) c -= 1;
unsigned int result = (input & nbitsmask) >> c;
I assumed you meant using only integer math.
I used some code from #OliCharlesworth's link, you could remove the conditionals too by using the LUT for trailing zeroes code there.

Get signed integer from 2 16-bit signed bytes?

So this sensor I have returns a signed value between -500-500 by returning two (high and low) signed bytes. How can I use these to figure out what the actual value is? I know I need to do 2's complement, but I'm not sure how. This is what I have now -
real_velocity = temp.values[0];
if(temp.values[1] != -1)
real_velocity += temp.values[1];
//if high byte > 1, negative number - take 2's compliment
if(temp.values[1] > 1) {
real_velocity = ~real_velocity;
real_velocity += 1;
}
But it just returns the negative value of what would be a positive. So for instance, -200 returns bytes 255 (high) and 56(low). Added these are 311. But when I run the above code it tells me -311. Thank you for any help.
-200 in hex is 0xFF38,
you're getting two bytes 0xFF and 0x38,
converting these back to decimal you might recognise them
0xFF = 255,
0x38 = 56
your sensor is not returning 2 signed bytes but a simply the high and low byte of a signed 16 bit number.
so your result is
value = (highbyte << 8) + lowbyte
value being a 16 bit signed variable.
Based on the example you gave, it appears that the value is already 2's complement. You just need to shift the high byte left 8 bits and OR the values together.
real_velocity = (short) (temp.values[0] | (temp.values[1] << 8));
You can shift the bits and mask the values.
int main()
{
char data[2];
data[0] = 0xFF; //high
data[1] = 56; //low
int value = 0;
if (data[0] & 0x80) //sign
value = 0xFFFF8000;
value |= ((data[0] & 0x7F) << 8) | data[1];
std::cout<<std::hex<<value<<std::endl;
std::cout<<std::dec<<value<<std::endl;
std::cin.get();
}
Output:
ffffff38
-200
real_velocity = temp.values[0];
real_velocity = real_velocity << 8;
real_velocity |= temp.values[1];
// And, assuming 32-bit integers
real_velocity <<= 16;
real_velocity >>= 16;
For 8-bit bytes, first just convert to unsigned:
typedef unsigned char Byte;
unsigned const u = (Byte( temp.values[1] ) << 8) | Byte( temp.values[0] );
Then if that is greater than the upper range for 16-bit two's complement, subtract 216:
int const i = int(u >= (1u << 15)? u - (1u << 16) : u);
You could do tricks at the bit level, but I don't think there's any point in that.
The above assuming that CHAR_BIT = 8, that unsigned is more than 16 bits, and that the machine and desired result is two's complement.
#include <iostream>
using namespace std;
int main()
{
typedef unsigned char Byte;
struct { char values[2]; } const temp = { 56, 255 };
unsigned const u = (Byte( temp.values[1] ) << 8) | Byte( temp.values[0] );
int const i = int(u >= (1u << 15)? u - (1u << 16) : u);
cout << i << endl;
}

How to read individual bits from an array?

Lets say i have an array dynamically allocated.
int* array=new int[10]
That is 10*4=40 bytes or 10*32=320 bits. I want to read the 2nd bit of the 30th byte or 242nd bit. What is the easiest way to do so? I know I can access the 30th byte using array[30] but accessing individual bits is more tricky.
bool bitset(void const * data, int bitindex) {
int byte = bitindex / 8;
int bit = bitindex % 8;
unsigned char const * u = (unsigned char const *) data;
return (u[byte] & (1<<bit)) != 0;
}
this is working !
#define GET_BIT(p, n) ((((unsigned char *)p)[n/8] >> (n%8)) & 0x01)
int main()
{
int myArray[2] = { 0xaaaaaaaa, 0x00ff00ff };
for( int i =0 ; i < 2*32 ; i++ )
printf("%d", GET_BIT(myArray, i));
return 0;
}
ouput :
0101010101010101010101010101010111111111000000001111111100000000
Be carefull of the endiannes !
First, if you're doing bitwise operations, it's usually
preferable to make the elements an unsigned integral type
(although in this case, it really doesn't make that much
difference). As for accessing the bits: to access bit i in an
array of n int's:
static int const bitsPerWord = sizeof(int) * CHAR_BIT;
assert( i >= 0 && i < n * bitsPerWord );
int wordIndex = i / bitsPerWord;
int bitIndex = i % bitsPerWord;
then to read:
return (array[wordIndex] & (1 << bitIndex)) != 0;
to set:
array[wordIndex] |= 1 << bitIndex;
and to reset:
array[wordIndex] &= ~(1 << bitIndex);
Or you can use bitset, if n is constant, or vector<bool> or
boost::dynamic_bitset if it's not, and let someone else do the
work.
You can use something like this:
!((array[30] & 2) == 0)
array[30] is the integer.
& 2 is an and operation which masks the second bit (2 = 00000010)
== 0 will check if the mask result is 0
! will negate that result, because we're checking if it's 1 not zero....
You need bit operations here...
if(array[5] & 0x1)
{
//the first bit in array[5] is 1
}
else
{
//the first bit is 0
}
if(array[5] & 0x8)
{
//the 4th bit in array[5] is 1
}
else
{
//the 4th bit is 0
}
0x8 is 00001000 in binary. Doing the anding masks all other bits and allows you to see if the bit is 1 or 0.
int is typically 32 bits, so you would need to do some arithmetic to get a certain bit number in the entire array.
EDITED based on comment below - array contains int of 32 bits, not 8 bits uchar.
int pos = 241; // I start at index 0
bool bit242 = (array[pos/32] >> (pos%32)) & 1;