This code is from a problem in HackerRank (Q: Printing Tokens in C).

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

int main() {
    char *s = malloc(1024 * sizeof(char));  // Allocate for string
    scanf("%[^\n]", s);
    s = realloc(s, strlen(s)+1);   // Here's my doubt

    // Solution
    
    return 0;
}

Forget about the question, look at the following iteration-

s = realloc( s, strlen(s)+1 );   // +1

This line of code reallocates the memory of the string ‘s’ according to the length of the string ‘s’, It also increases the length for a null terminator ‘\0’. But strlen() only measures until it reaches ‘\0’, but there’s no ‘\0’ in ‘s’ as we’re adding a byte for ‘\0’. How can we assume it will measure the perfect length of string ‘s’? If it measures the perfect length of ‘s’ then there must be a null terminator at the end so why add 1 again?

Someone explain this, there’s no null terminator ‘\0’ so we add a byte for it.

Or Am I completely wrong about this?

>Solution :

Someone explain this, there’s no null terminator ‘\0’ so we add a byte for it.

You are misinterpreting what happens here.

scanf is used to read a string into the memory pointed to by s.
In C, a string is a nul-terminated sequence of characters. That means, it is responsibility of scanf to place the 0 byte there after the last character from the input.

If that weren’t the case, you would not be allowed to pass s to strlen at all because it expects a nul-terminated string.

In realloc the +1 is not added to add a 0 byte. It is added to keep the memory of that byte. You start with a large buffer and resize to the length you need. For this, you must add 1 to prevent the last byte from being freed during realloc.

So, the assumption that there would not be a nul-terminator, is wrong. It is there all the time and shall still be there after reducing the buffer size.