Not dereferencing produces a warning when compiling

November 16, 2023

I’m struggling to understand the nuances of char arrays and string literals, and I’m quite confused as to why the following code produces a warning when I compile it.

warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[4]’ [-Wformat=]

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

int main(){
    char str_arr[4] = "abc";
    char (*p)[4];
    p = &str_arr;
    printf("A single character: %c\n",(*p)[1]);  
    printf("The entire string: %s\n",p); // When I replace p with *p, the warning goes away, why?
    return 0;
}

If I understand correctly, p is a pointer to a char array of size 4 and after assigning it, it now points to the char array str_array. If we dereference p that gives us the entire char array, but doesn’t %s require a char* as an argument? If we have a char array in place of a char pointer in the argument, shouldn’t that produce a warning?

>Solution :

The format %s expects an argument of type char *. The type of p is not char *. But when you dereference p, with *p, you get an array of char that will decay to char *.

Some compilers are able to detect format-string and argument mismatch and warn about it, because it will lead to undefined behavior.