Samashki95 - 1 month ago 6

C Question

I'm implementing a JPEG-Decoder and in one step I need to determine the sign of a value from a given number of bits

`(positive if 1th bit = 1)`

When the 1th bit is

`0`

I've have the following function to do this job:

`#include <stdio.h> /* printf */`

#include <string.h> /* strcat */

#include <stdlib.h> /* strtol */

typedef int bool;

#define true 1

#define false 0

int DetermineSign(int val, int nBits)

{

bool negative = val < (1<<(nBits-1));

if (negative)

{

// (-1 << (s)), makes the last bit a 1, so we have 1000,0000 for example for 8 bits

val = val + (-1 << (nBits)) + 1;

}

// Else its unsigned, just return

return val;

}

Could anyone explain please what does this expression

`(-1 << (nBits))`

I know there is a comment from author to explain it, but I also tested it with the following function and it returns another result.

`const char *byte_to_binary(int x)`

{

static char b[9];

b[0] = '\0';

int z;

for (z = 128; z > 0; z >>= 1)

{

strcat(b, ((x & z) == z) ? "1" : "0");

}

return b;

}

int main(void)

{

char testValue = 0;

testValue = (-1 <<(testValue));

printf("%s\n", byte_to_binary(testValue)); // output 1111 1111 doesn't it has to be 1000 000?

return 0;

}

Thank you!

Answer

It replaces the right-most (least significant) ones with zeros:

`-1 << 0 == 0xFFFFFFFF`

`-1 << 1 == 0xFFFFFFFE`

`-1 << 2 == 0xFFFFFFFC`

`-1 << 3 == 0xFFFFFFF8`

`-1 << 4 == 0xFFFFFFF0`

- ...

Source (Stackoverflow)

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