Question

#include <stdio.h>
int main(void)
{
    int i,j,k;
    char st;
    printf("enter string\n");
    scanf("%s", st);
    printf("the entered string is %s\n", st);
}

Compiling above program gives me a warning:

warning: format '%s' expects argument of type 'char *', but argument 2 has type 'int' [-Wformat]
palindrom.c:8:1: warning: format '%s' expects argument of type 'char *', but argument 2 has type 'int' [-Wformat]

What am I doing wrong here?

This is what happens when I run it:

$ ./a.out
enter string
kiaaa
the entered string is (null)

Edit:

Here is another version of the code (made char st; into char *st):

#include <stdio.h>
int main(void)
{
    int i,j,k;
    char *st;
    printf("enter string\n");
    scanf("%s", st);
    printf("the entered string is %s\n", st);
}

However, it behaves the same on runtime.

Answer 1

char st is a single character. Judging by the rest of your code, you probably intended to declare an array of characters:

char st[80];