I wonder why &a(Data type: char) is printed as value not address, using %s

Question

#include <stdio.h>

int main(void)
{

    char var = 'z';

    printf("[buf]: %s \n", &var); // the output is z~~~~~, but the first char of output is only z. why??
}

Answer 1

The conversion specifier s is designed to output strings (or their parts): sequences of characters terminated by the zero character '\0'.

To output the address of an object there is the conversion specifier p.

Here is a demonstrative program.

#include <stdio.h>

int main(void) 
{
    char s[] = "Hello";
    
    printf( "%s\n", s );
    printf( "%p\n", ( void * )s );
    
    return 0;
}

The program output might look like

Hello
0x7ffc86fe4372

As for the code in your question then it has undefined behavior because the expression &var does not point to a string because the variable var is defined like

char var = 'z';

If you want to output its address then you can do it as

printf("[buf]: %p \n", ( void * )&var);

Answer 2

%s tells printf to accept a pointer to the first character of a string and to print that string, up to the null character that indicates its end. Since you pass the address of a single character, printf prints that and continues looking in memory for more characters to print, until it finds a byte containing zero. For %s, when you pass a pointer to a single character, rather than an array of characters terminated by a null character, the behavior is not defined by the C standard.

To print an address, use %p and convert the pointer to void *:

printf("%p\n", (void *) &var);