What would be the fastest way possible to reverse the nibbles (e.g digits) of a hexadecimal number in C++?
Here's an example of what I mean : 0x12345 -> 0x54321
Here's what I already have:
unsigned int rotation (unsigned int hex) {
unsigned int result = 0;
while (hex) {
result = (result << 4) | (hex & 0xF);
hex >>= 4;
}
return result;
}
This problem can be split into two parts:
Reverse the nibbles of an integer. Reverse the bytes, and swap the nibble within each byte.
Shift the reversed result right by some amount to adjust for the "variable length". There are std::countl_zero(x) & -4 (number of leading zeroes, rounded down to a multiple of 4) leading zero bits that are part of the leading zeroes in hexadecimal, shifting right by that amount makes them not participate in the reversal.
For example, using some of the new functions from <bit>:
#include <stdint.h>
#include <bit>
uint32_t reverse_nibbles(uint32_t x) {
// reverse bytes
uint32_t r = std::byteswap(x);
// swap adjacent nibbles
r = ((r & 0x0F0F0F0F) << 4) | ((r >> 4) & 0x0F0F0F0F);
// adjust for variable-length of input
int len_of_zero_prefix = std::countl_zero(x) & -4;
return r >> len_of_zero_prefix;
}
That requires C++23 for std::byteswap which may be a bit optimistic, you can substitute it with some other byteswap.
Easily adaptable to uint64_t too.
i would do it without loops based on the assumption that the input is 32 bits
result = (hex & 0x0000000f) << 28
| (hex & 0x000000f0) << 20
| (hex & 0x00000f00) << 12
....
dont know if faster, but I find it more readable
I'd like to have (1st bit + 2nd bit) in decimal number.
unsigned char a = 52;
unsigned char b = ((((a >> 2) & 0x1)*2)+(((a >> 1) & 0x1)*1));
// expected output: 2
My code works, but I think this way is pretty bad.
I think the confusion is in your phrasing... You say you want 1st bit + 2nd bit, which I would argue (semantically) would be bit index 0 and 1. But from your code it's clear you really want bit index 1 and 2, hence why 52 (110100) would yield 2.
In this case the code you provide is easily simplified by shifting the bits one to the right and grabbing the low 2 bits:
unsigned char a = 52;
unsigned char b = (a >> 1) & 3;
Firstly, if anyone has a better title for me, let me know.
Here is an example of the process I am trying to automate with C++
I have an array of values that appear in this format:
9C07 9385 9BC7 00 9BC3 9BC7 9385
I need to convert them to binary and then convert every 5 bits to decimal like so with the last bit being a flag:
I'll do this with only the first word here.
9C07
10011 | 10000 | 00011 | 1
19 | 16 | 3
These are actually x,y,z coordinates and the final bit determines the order they are in a '0' would make it x=19 y=16 z=3 and '1' is x=16 y=3 z=19
I already have a buffer filled with these hex values, but I have no idea where to go from here.
I assume these are integer literals, not strings?
The way to do this is with bitwise right shift (>>) and bitwise AND (&)
#include <cstdint>
struct Coordinate {
std::uint8_t x;
std::uint8_t y;
std::uint8_t z;
constexpr Coordinate(std::uint16_t n) noexcept
{
if (n & 1) { // flag
x = (n >> 6) & 0x1F; // 1 1111
y = (n >> 1) & 0x1F;
z = n >> 11;
} else {
x = n >> 11;
y = (n >> 6) & 0x1F;
z = (n >> 1) & 0x1F;
}
}
};
The following code would extract the three coordinates and the flag from the 16 least significant bits of value (ie. its least significant word).
int flag = value & 1; // keep only the least significant bit
value >>= 1; // shift right by one bit
int third_integer = value & 0x1f; // keep only the five least significant bits
value >>= 5; // shift right by five bits
int second_integer = value & 0x1f; // keep only the five least significant bits
value >>= 5; // shift right by five bits
int first_integer = value & 0x1f; // keep only the five least significant bits
value >>= 5; // shift right by five bits (only useful if there are other words in "value")
What you need is most likely some loop doing this on each word of your array.
This question is asked on Pearls of programming Question 2. And I am having trouble understanding its solution.
Here is the solution written in the book.
#define BITSPERWORD 32
#define SHIFT 5
#define MASK 0x1F
#define N 10000000
int a[1 + N/BITSPERWORD];
void set(int i) { a[i>>SHIFT] |= (1<<(i & MASK)); }
void clr(int i) { a[i>>SHIFT]&=~(1<<(i & MASK)); }
int test(int i) { return a[i>>SHIFT]&(1<<(i & MASK)); }
I have ran this in my compiler and I have looked at another question that talks about this problem, but I still dont understand how this solution works.
Why does it do a[i>>SHIFT]? Why cant it just be a[i]=1; Why does i need to shifted right 5 times?
32 is 25, so a right-shift of 5 bits is equivalent to dividing by 32. So by doing a[i>>5], you are dividing i by 32 to figure out which element of the array contains bit i -- there are 32 bits per element.
Meanwhile & MASK is equivalent to mod 32, so 1<<(i & MASK) builds a 1-bit mask for the particular bit within the word.
Divide the 32 bits of int i (starting form bit 0 to bit 31) into two parts.
First part is the most significant bits 31 to 5. Use this part to find the index in the array of ints (called a[] here) that you are using to implement the bit array. Initially, the entire array of ints is zeroed out.
Since every int in a[] is 32 bits, it can keep track of 32 ints with those 32 bits. We divide every input i with 32 to find the int in a[] that is supposed to keep track of this i.
Every time a number is divided by 2, it is effectively right shifted once. To divide a number by 32, you simply right shift it 5 times. And that is exactly what we get by filtering out the first part.
Second part is the least significant bits 0 to 4. After a number has been binned into the correct index, use this part to set the specific bit of the zero stored in a[] at this index. Obviously, if some bit of the zero at this index has already been set, the value at that index will not be zero anymore.
How to get the first part? Right shifting i by 5 (i.e. i >> SHIFT).
How to get the second part? Do bitwise AND of i by 11111. (11111)2 = 0x1F, defined as MASK. So, i & MASK will give the integer value represented by the last 5 bits of i.
The last 5 bits tell you how many bits to go inside the number in a[]. For example, if i is 5, you want to set the bit in the index 0 of a[] and you specifically want to set the 5th bit of the int value a[0].
Index to set = 5 / 32 = (0101 >> 5) = 0000 = 0.
Bit to set = 5th bit inside a[0]
= a[0] & (1 << 5)
= a[0] & (1 << (00101 & 11111)).
Setting the bit for given i
Get the int to set by a[i >> 5]
Get the bit to set by pushing a 1 a total of i % 32 times to the left i.e. 1 << (i & 0x1F)
Simply set the bit as a[i >> 5] = a[i >> 5] | (1 << (i & 0x1F));
That can be shortened to a[i >> 5] |= (1 << (i & 0x1F));
Getting/Testing the bit for given i
Get the int where the desired bit lies by a[i >> 5]
Generate a number where all bits except for the i & 0x1F bit are 0. You can do that by negating 1 << (i & 0x1F).
AND the number generated above with the value stored at this index in a[]. If the value is 0, this particular bit was 0. If the value is non-zero, this bit was 1.
In code you would simply, return a[i >> 5] & (1 << (i & 0x1F)) != 0;
Clearing the bit for given i: It means setting the bit for that i to 0.
Get the int where the bit lies by a[i >> 5]
Get the bit by 1 << (i & 0x1F)
Invert all the bits of 1 << (i & 0x1F) so that the i's bit is 0.
AND the number at this index and the number generated in step 3. That will clear i's bit, leaving all other bits intact.
In code, this would be: a[i >> 5] &= ~(1 << (i & 0x1F));
Assuming I have a byte b with the binary value of 11111111
How do I for example read a 3 bit integer value starting at the second bit or write a four bit integer value starting at the fifth bit?
Some 2+ years after I asked this question I'd like to explain it the way I'd want it explained back when I was still a complete newb and would be most beneficial to people who want to understand the process.
First of all, forget the "11111111" example value, which is not really all that suited for the visual explanation of the process. So let the initial value be 10111011 (187 decimal) which will be a little more illustrative of the process.
1 - how to read a 3 bit value starting from the second bit:
___ <- those 3 bits
10111011
The value is 101, or 5 in decimal, there are 2 possible ways to get it:
mask and shift
In this approach, the needed bits are first masked with the value 00001110 (14 decimal) after which it is shifted in place:
___
10111011 AND
00001110 =
00001010 >> 1 =
___
00000101
The expression for this would be: (value & 14) >> 1
shift and mask
This approach is similar, but the order of operations is reversed, meaning the original value is shifted and then masked with 00000111 (7) to only leave the last 3 bits:
___
10111011 >> 1
___
01011101 AND
00000111
00000101
The expression for this would be: (value >> 1) & 7
Both approaches involve the same amount of complexity, and therefore will not differ in performance.
2 - how to write a 3 bit value starting from the second bit:
In this case, the initial value is known, and when this is the case in code, you may be able to come up with a way to set the known value to another known value which uses less operations, but in reality this is rarely the case, most of the time the code will know neither the initial value, nor the one which is to be written.
This means that in order for the new value to be successfully "spliced" into byte, the target bits must be set to zero, after which the shifted value is "spliced" in place, which is the first step:
___
10111011 AND
11110001 (241) =
10110001 (masked original value)
The second step is to shift the value we want to write in the 3 bits, say we want to change that from 101 (5) to 110 (6)
___
00000110 << 1 =
___
00001100 (shifted "splice" value)
The third and final step is to splice the masked original value with the shifted "splice" value:
10110001 OR
00001100 =
___
10111101
The expression for the whole process would be: (value & 241) | (6 << 1)
Bonus - how to generate the read and write masks:
Naturally, using a binary to decimal converter is far from elegant, especially in the case of 32 and 64 bit containers - decimal values get crazy big. It is possible to easily generate the masks with expressions, which the compiler can efficiently resolve during compilation:
read mask for "mask and shift": ((1 << fieldLength) - 1) << (fieldIndex - 1), assuming that the index at the first bit is 1 (not zero)
read mask for "shift and mask": (1 << fieldLength) - 1 (index does not play a role here since it is always shifted to the first bit
write mask : just invert the "mask and shift" mask expression with the ~ operator
How does it work (with the 3bit field beginning at the second bit from the examples above)?
00000001 << 3
00001000 - 1
00000111 << 1
00001110 ~ (read mask)
11110001 (write mask)
The same examples apply to wider integers and arbitrary bit width and position of the fields, with the shift and mask values varying accordingly.
Also note that the examples assume unsigned integer, which is what you want to use in order to use integers as portable bit-field alternative (regular bit-fields are in no way guaranteed by the standard to be portable), both left and right shift insert a padding 0, which is not the case with right shifting a signed integer.
Even easier:
Using this set of macros (but only in C++ since it relies on the generation of member functions):
#define GETMASK(index, size) ((((size_t)1 << (size)) - 1) << (index))
#define READFROM(data, index, size) (((data) & GETMASK((index), (size))) >> (index))
#define WRITETO(data, index, size, value) ((data) = (((data) & (~GETMASK((index), (size)))) | (((value) << (index)) & (GETMASK((index), (size))))))
#define FIELD(data, name, index, size) \
inline decltype(data) name() const { return READFROM(data, index, size); } \
inline void set_##name(decltype(data) value) { WRITETO(data, index, size, value); }
You could go for something as simple as:
struct A {
uint bitData;
FIELD(bitData, one, 0, 1)
FIELD(bitData, two, 1, 2)
};
And have the bit fields implemented as properties you can easily access:
A a;
a.set_two(3);
cout << a.two();
Replace decltype with gcc's typeof pre-C++11.
You need to shift and mask the value, so for example...
If you want to read the first two bits, you just need to mask them off like so:
int value = input & 0x3;
If you want to offset it you need to shift right N bits and then mask off the bits you want:
int value = (intput >> 1) & 0x3;
To read three bits like you asked in your question.
int value = (input >> 1) & 0x7;
just use this and feelfree:
#define BitVal(data,y) ( (data>>y) & 1) /** Return Data.Y value **/
#define SetBit(data,y) data |= (1 << y) /** Set Data.Y to 1 **/
#define ClearBit(data,y) data &= ~(1 << y) /** Clear Data.Y to 0 **/
#define TogleBit(data,y) (data ^=BitVal(y)) /** Togle Data.Y value **/
#define Togle(data) (data =~data ) /** Togle Data value **/
for example:
uint8_t number = 0x05; //0b00000101
uint8_t bit_2 = BitVal(number,2); // bit_2 = 1
uint8_t bit_1 = BitVal(number,1); // bit_1 = 0
SetBit(number,1); // number = 0x07 => 0b00000111
ClearBit(number,2); // number =0x03 => 0b0000011
You have to do a shift and mask (AND) operation.
Let b be any byte and p be the index (>= 0) of the bit from which you want to take n bits (>= 1).
First you have to shift right b by p times:
x = b >> p;
Second you have to mask the result with n ones:
mask = (1 << n) - 1;
y = x & mask;
You can put everything in a macro:
#define TAKE_N_BITS_FROM(b, p, n) ((b) >> (p)) & ((1 << (n)) - 1)
"How do I for example read a 3 bit integer value starting at the second bit?"
int number = // whatever;
uint8_t val; // uint8_t is the smallest data type capable of holding 3 bits
val = (number & (1 << 2 | 1 << 3 | 1 << 4)) >> 2;
(I assumed that "second bit" is bit #2, i. e. the third bit really.)
To read bytes use std::bitset
const int bits_in_byte = 8;
char myChar = 's';
cout << bitset<sizeof(myChar) * bits_in_byte>(myChar);
To write you need to use bit-wise operators such as & ^ | & << >>. make sure to learn what they do.
For example to have 00100100 you need to set the first bit to 1, and shift it with the << >> operators 5 times. if you want to continue writing you just continue to set the first bit and shift it. it's very much like an old typewriter: you write, and shift the paper.
For 00100100: set the first bit to 1, shift 5 times, set the first bit to 1, and shift 2 times:
const int bits_in_byte = 8;
char myChar = 0;
myChar = myChar | (0x1 << 5 | 0x1 << 2);
cout << bitset<sizeof(myChar) * bits_in_byte>(myChar);
int x = 0xFF; //your number - 11111111
How do I for example read a 3 bit integer value starting at the second bit
int y = x & ( 0x7 << 2 ) // 0x7 is 111
// and you shift it 2 to the left
If you keep grabbing bits from your data, you might want to use a bitfield. You'll just have to set up a struct and load it with only ones and zeroes:
struct bitfield{
unsigned int bit : 1
}
struct bitfield *bitstream;
then later on load it like this (replacing char with int or whatever data you are loading):
long int i;
int j, k;
unsigned char c, d;
bitstream=malloc(sizeof(struct bitfield)*charstreamlength*sizeof(char));
for (i=0; i<charstreamlength; i++){
c=charstream[i];
for(j=0; j < sizeof(char)*8; j++){
d=c;
d=d>>(sizeof(char)*8-j-1);
d=d<<(sizeof(char)*8-1);
k=d;
if(k==0){
bitstream[sizeof(char)*8*i + j].bit=0;
}else{
bitstream[sizeof(char)*8*i + j].bit=1;
}
}
}
Then access elements:
bitstream[bitpointer].bit=...
or
...=bitstream[bitpointer].bit
All of this is assuming are working on i86/64, not arm, since arm can be big or little endian.