How can I print the last digit of INT_MIN integer using putchar only?-CodePudding

I am not allowed to use abs or any other libraries. Here is what I've tried so far.

#include <stdio.h>
#include <limits.h>
int main()
{
    int n = INT_MIN;

    if (n == INT_MIN)
    {
        int a = -n;
        a = n % 10;
        putchar(a   '0');
    }
    return (0);
}

The output is always "(" and not the correct answer.

CodePudding user response：

Apply modulus to the negative number first to get a single negative digit. Negate that.

#include <stdio.h>
#include <limits.h>
int main()
{
    int n = INT_MIN;

    int a = -( n % 10);
    putchar(a   '0');
    return (0);
}

CodePudding user response：

When n == INT_MIN, do not use a = -n as that is undefined behavior (UB) since -n overflows int math.

Using a = n will result in some value [-9 ... 0] since n < 0. No need to negate a before using it to print the digit, just subtract.

    int n = INT_MIN;
    int a = n;
    putchar('0' - a);

To print the entire int, note that there are more negative values [INT_MIN ... -1] then positive ones [1 ... INT_MAX]^*1. Instead of negating negative numbers to positives (and have UB with INT_MIN), negate positive ones to the negative side.

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

int n = {some value from INT_MIN to INT_MAX};
int a = n < 0 ? n : -n;

char buf[21];
char *p = buf   sizeof buf - 1;
*p = '\0';

do {
  p--;
  *p = '0' - a;
  a /= 10;
} while (a);

if (n < 0) {
  *--p = '-';
}
puts(p);

This approach works even if instead of int, we used long long or intmax_t. We just need to adjust the buffer length. char buf[21] big enough for 64-bit integers.

^*1 With the common 2's complement integer encoding, there is always one more negative. With the rare 1s' complement and sign-magnitude encoding, likely to be dropped in C2x, there are the same. Of course with those there may be a 0 and -0, yet let us leave that for historic programming.