Why I can't see that last char in array is null terminator?

Question

How can I check that a char array is null terminated?

char input[80];

fgets(input, sizeof(input), stdin);

for (i = 0; input[i]; i++)
{
    if (input[i] == '\0') {
       printf("null!!!"); // never works
    }
}

For example, the below code does not print null.

Answer 1

In your for loop condition, you're already checking for the non-0 value of input[i], so the inside if is a dead condition.

To test, make an infinite loop, check if and then, print and break out. Something like

for (i = 0; ; i++)
{
    if (input[i] == '\0') {
       printf("null!!!\n");  // now works
       break;
    }
}

Answer 2

As soon as input[i] is 0 the loop exits, so that print statement is never executed.

Answer 3

Here is another working code, if you are interested :

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

int main() {
    char input[] = "I am a string !!";
    int i = 0;
    while(1) {
        if (input[i] == '\0') {
           printf("null!!!");
           break;
        }
        i++;
    }
}

Answer 4

All answers here don't check memory bounds, so you'll definitely run into a segmentation fault. Still, if an error occurs during freading, you may have garbage in input.

char input[80];

fgets(input, sizeof(input), stdin);

for (i = 0; i<80; i++)
{
    if (input[i] == '\0') {
        printf("null!!!"); 
    }
}

You could also change if (input[i] == '\0') to just if (! input[i]) since '\0' (null-terminator) has ASCII value 0.