Search common characters in strings and return pointer to the common character (without using strpbrk)

Question

Following code finds the common character in two strings and returns a pointer to the common string if any. I'm trying to emulate the function strpbrk without using any builtin functions or subscripts.

#include <stdio.h>
#include <stdlib.h>



char *find_char( char const *source, char const *chars ){

    char* str1;
    char* str2;

    if (source == NULL || chars == NULL)
        return NULL;

     else {

        for( *str1=&source != '\0'; ++str1;){
            for( *str2=&chars != '\0'; ++str2;){

                if (*str1 == *str2)

                    return str1;

                else return NULL;}

    }

}
}

char* main(){  
    char const *source = "ab";
    char const *chars = "bc";
    find_char(source, chars);
}

But I'm getting the following error

Running "/home/ubuntu/workspace/hello-c-world.c"
bash: line 12: 29755 Segmentation fault "$file.o" $args
Process exited with code: 139

I am a beginner at C and learning how to manipulate pointers, please let me know what am I doing wrong and how I can strengthen my C programming skills, right now I'm mostly using Kenneth Reek's book "Pointers on C"

Thanks

Answer 1

Various ìf logic and runaway pointer problems. Fixed code below:

char *find_char( char const *source, char const *chars ){

    char* str1;
    char* str2;

    if (source != NULL && chars != NULL) {
       for( str1=source; *str1; str1++){
           for( str2=chars; *str2; str2++){
               if (*str1 == *str2) return str1;
           }
       }
    }
    return NULL;
}

char* main(){  
    char const *source = "ab";
    char const *chars = "bc";
    char *x;
    x = find_char(source, chars);
    if (x != NULL) printf("%c", *x);
}