Why does program crash for dereferencing pointer to char by using '%s'?

Question

I am learning c, a beginner, can anybody please make me understood which concept am I missing? And thanks in advance.

#include<stdio.h>

int main()
{
    char s[10];
    s[0]='A';
    s[1]='B';
    s[2]='\0';
    char *p;
    int i;

    p=s;


    printf("%c\n", *p); //It's ok.


    printf("%s", *p); // or *s...what's wrong here,why does program crash?

    return 0;
}

Answer 1

Change

printf("%s", *p);

to

printf("%s", p);

The reason why is that %s is expecting a pointer, and *p is the dereferenced value at p , aka the char value at p[0] .

If this doesn't make sense, picture why printf("%c\\n", *p) works. *p is the same as p[0] , which is the same as s[0] since p points to s . Because s[0] is a char , %c works here because it is expecting a char . But %s on the other hand expects char * .

Answer 2

You want printf("%s", p) . Don't dereference the pointer.