Why NULL terminated string's length is smaller than expected in C?

There is code:

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

int main()
{
  char *str = calloc(3, sizeof(char));
  str[2] = '\0';
  
  printf("%ld", strlen(str));

  free(str);
}

Expected output 2
Actual output 0

Why?

As far as I understood strlen counts symbols before \0. Maybe str[0] and str[1] are already ‘\0’?

>Solution :

calloc() initializes allocated memory with all \0. You then set the last byte to \0 – it’s already that value from initialization.

strlen() counts the chars before the first \0. Since the very first char is already \0, the length is zero.