Absolute value of INT_MIN

Question

How could I extract the absolute value of INT_MIN without overflowing? See this code for the problem:

#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

int main(void) {
    printf("INT_MAX: %d\n", INT_MAX);
    printf("INT_MIN: %d\n", INT_MIN);
    printf("abs(INT_MIN): %d\n", abs(INT_MIN));

    return 0;
}

Spits out the following

INT_MAX: 2147483647
INT_MIN: -2147483648
abs(INT_MIN): -2147483648

I need this for a check if an int value is greater than zero.

As for this question being a duplicate of Why the absolute value of the max negative integer -2147483648 is still -2147483648?, I have to disagree, since this is a HOW, not a WHY question.

Answer 1

The %d conversion specifier in the format string of printf converts the corresponding argument to a signed decimal integer, which in this case, overflows for the int type. C standard specifically mentions that signed integer overflow is undefined behaviour. What you should do is to use %u in the format string. Also, you need to include the headers stdio.h and stdlib.h for the prototype of the functions printf and abs respectively.

#include <limits.h>
#include <stdio.h>
#include <stdlib.h>

// This solves the issue of using the standard abs() function
unsigned int absu(int value) {
    return (value < 0) ? -((unsigned int)value) : (unsigned int)value;
}

int main(void) {
    printf("INT_MAX: %d\n", INT_MAX);
    printf("INT_MIN: %d\n", INT_MIN);
    printf("absu(INT_MIN): %u\n", absu(INT_MIN));

    return 0;
}

This gives the output on my 32-bit machine:

INT_MAX: 2147483647
INT_MIN: -2147483648
absu(INT_MIN): 2147483648