This month’s Exercise may not have been challenging. Still, remember that blog is about learning the C language. Sometimes performing what could be a simple task may end up being complex than you anticipated.

Sadly, this challenge probably wasn’t that complex. Still, it opens an insight into a programming quirk you may not have expected. Before revealing the quirk, here is my solution:

2024_04-Exercise.c

#include <stdio.h>

int main()
{
    int a;

    printf("Enter an integer: ");
    scanf("%d",&a);

    if( a<0 )
        a = -a;

    printf("The absolute value is %d\n",a);

    return 0;
}

The code prompts for an integer value, int a. If the value is less than zero — meaning that it’s negative — the negation operator flips the value positive: a = -a. This expression isn’t the same as a -= a, which would always be zero. No, it negates the negative, turning it positive. The result is output.

Here’s a sample run:

Enter an integer: -200
The absolute value is 200

And now for the quirk:

Enter an integer: -2147483648
The absolute value is -2147483648

When a negative value is input at the bottom of the signed integer’s range, the result is the same negative value.

The range for a 32-bit signed int data type in C is from -2,147,483,648 to 2,147,483,647. Due to the way negative values are expressed in binary, you cannot flip the lowest negative value to a positive as it would be out of range: The upper limit of an int data type is 2,147,483,647, not 2,147,483,648.

This effect holds for the C library’s abs() function as well: If you recode the solution to use abs() (which requires including the stdlib.h header file), you get the same results.

Here’s the nerdy part: When incremented, the 32-bit value 2,147,483,647 becomes -2,147,483,648. Therefore, the absolute value of -2,147,483,648 cannot be calculated as it’s beyond the range of a signed 32-bit integer.

I hope you enjoyed the challenge and their weird aspect of finding the absolute value of an integer’s negative minimum.