Samashki95 - 1 year ago 130
C Question

# Bit- manipulation

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`
then I have to get the two's complement from this value and additionaly add 1 to result.

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))`
do and how it works?
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!

• `-1 << 0 == 0xFFFFFFFF`
• `-1 << 1 == 0xFFFFFFFE`
• `-1 << 2 == 0xFFFFFFFC`
• `-1 << 3 == 0xFFFFFFF8`
• `-1 << 4 == 0xFFFFFFF0`