The largest negative number in 2's complement

[johnmerlino]

In this exercise:

The C Programming Language Exercise 3-4

It states the following:

"In a two's complement number representation, our version of itoa does not handle the largest negative number, that is, the value of n equal to -(2 to the power (wordsize - 1)) ."

A char is one byte (255 bits). The range of an 8 bit variable using a two's complement representation is -128 to 127. Therefore -128 is the largest negative value. The statement in book suggests that the itoa function will not output -128 if we pass -128 as a parameter, because in itoa when we try to convert -128 to positive -128, the inverse of -128 is -128. However, I just ran this code in my computer and it successfully outputted -128. Am I missing something?

#include <stdio.h>
#include <string.h>

#define SIZE 10

void reverse(char s[])
{
    int c, i, j;

    for (i = 0, j = strlen(s)-1; i < j; i++, j--) {
        c = s[i];
        s[i] = s[j];
        s[j] = c;
    }
}

void itoa(int n, char s[])
{
    int i, sign;

    if ((sign = n) < 0)
        n = -n;

    i = 0;

    do {
        s[i++] = n % 10 + '0';

    } while ((n /= 10) > 0);


    if (sign < 0)
        s[i++] = '-';

    s[i] = '\0';

    reverse(s);
}

int main(void)
{
  char buffer[SIZE];
  itoa(-128, buffer);
  printf("Buffer: %s\n", buffer);

  return 0;
}

[King Mir]

itoa takes an int as it's first argument, which is usually 32 bits. So presumably that version itoa will fail on -2,147,483,648 not -128.

[johnmerlino]

When I called the function with itoa(-2147483648, buffer);, I got a warning " warning: this decimal constant is unsigned only in ISO C90 [enabled by default]".