Testing for the presence of a character in an string in C

Question

What's wrong with this?

#include <stdio.h>

void main(){
    char *s="some text";
    printf("%d",is_in(s,'t'));

}

int is_in(char *s, char c){
    while(*s){
        if(*s==c) return 1;
        s++;
    }
    return 0;
}

I get the following compile time error with GCC:

test.c:9: error: conflicting types for ‘is_in’

test.c:9: note: an argument type that has a default promotion can’t match an empty parameter name list declaration

test.c:5: note: previous implicit declaration of ‘is_in’ was here

Answer 1

You are incrementing the character, not the pointer. Change *s++ to simply s++. In addition, you have forgotten to forward-declare your function "is_in". One other note: you should probably make your string a "const char*" instead of "char*", and, IMHO, explicit comparison with '\0' and using indexes is clearer:

#include <stdio.h>

int is_in(const char*, char);
int main(int argc, char* argv[]){
    const char* str="some text";
    printf("%d",is_in(s,'t'));
    return 0;
}

int is_in(const char* str, char c){
    int idx=0;
    while ( str[idx] != '\0' ){
        if ( str[idx] == c ){
            return 1;
        }
        idx++;
    }
    return 0;
}

Answer 2

Have you tried putting the is_in function above main?