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bitwise operator--complement

I want to use bitwise operator--complement on an integer array. Here is my code:

#include <stdio.h>
int number[100];
unsigned char temp[100],complement[100];

main()
{
number[1]=23;
/*type cast integer*/
temp[1]=(unsigned char)(number[1]);
/*use complement operator*/
complement[1]=~temp[1];
printf("u%",complement[1]);
return 0;
}

I didn't get the correct result,.Is there anything wrong?

Thanks, --km

homer99

kxm try this:
#include <stdio.h>
int number[100];
main()
{
number[1]=23;
printf("%d", ~number[1]);
}

pjknibbs

What result *are* you getting?

ozo

changing "u%" to "%u" gets the correct result 232

mahno

Ur program works correct:
23 = 00010111
232 = 11101000

fremsley

i guess 'u%' was just a typo.

complement[1] is of type unsigned char, so printf only pushes 8 significant bits as parameter on the stack; the '%u' tells printf to treat the parameter as an unsigned int (usually 32 significant bits). so try:

printf("%u", (unsigned int)(complement[i]));

Pacman

kxm, note that your code only converts the lo-byte of the integer.

C_Dreamer

Note further that you shouldn't really muck about with the sign bit of a signed int.

If you're going to twiddle bits, do it on unsigned types.

Here's a safe bit-twiddler:

unsigned int bitwiseNot(unsigned int n)
{
return ~n;
}

Wow, that was a tough one. :-)

ASKER CERTIFIED SOLUTION

balakrishnan_t

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C_Dreamer

An int is not necessarily two bytes. It's sizeof(int) bytes. Under Visual C++ 1.5, that gives you a value of 2. Under Visual C++ 5.0, it gives you a value of 4. On a Cray, sizeof(int) is 8.

Furthermore, bear in mind that bytes don't necessarily have 8 bits. They have /at least/ 8 bits. To find out how many bits per byte in your implementation, print out the value of CHAR_BIT (in <limits.h>). Of course, it's 99 chances out of 100 that it'll be 8! :-)

The problem with twiddling bits on an int is that you are assuming your representation is twos complement. On ones complement machines, you'll get a totally different result. Hence my earlier admonition.

I certainly agree that changing u% to %u is likely to result in more effective display of the result. <grin>

kxm

ASKER

Thanks