What is the behavior of scanf when using “%s” form

问题:

#include <stdio.h>

int main(void){

char a;
scanf("%s", &a);

return 0;
}

Works for me, but I wonder if in other context there might be any unexpected results.

回答1:

Take this code for example -

#include <stdio.h>

int main(void){

    char a;
    char b = '1';
    char c = '1';
    scanf("%s", &a);

    printf("%c = %c\n", b, c);
    return 0;
}

You would expect it to print 1 = 1, but just now when I ran it, it printed = 1 (at least in my compiler, don't expect anything stable from it)

scanf writes a string to the address of a, expecting enough space was allocated there (which is wrong in this case), this string has the input char and the null terminator. The null terminator overwrites some other memory, in my case - that of b. This is undefined behavior - don't do that (at least not while expecting it to make any sense).

回答2:

Either use format %c to match single char, or use %s to match a string.

Mixing them up (like in your code) is undefined behavior, anything may happen.