Printing Char Array Element

Question

I've been hung up on this for the past two hours, and it's really starting to irritate me. I'm using standard C, trying to print a char array's element.

The following is a snippet that works(prints entire array),

CreditCard validate_card(long long n) {

    CreditCard cc; // specify new credit card
    cc.n = n; // specify card num as passed
    cc.valid = false; // initialize as invalid
    cc.type = AEX; // initialize at american express

    bool valid;

    char s[20];
    sprintf( s, "%d", n ); // convert credit card number into char array
    printf("%s\n", s);
    return cc;
}

The following snippet does not work,

CreditCard validate_card(long long n) {

    CreditCard cc; // specify new credit card
    cc.n = n; // specify card num as passed
    cc.valid = false; // initialize as invalid
    cc.type = AEX; // initialize at american express

    bool valid;

    char s[20];
    sprintf( s, "%d", n ); // convert credit card number into char array
    printf("%s\n", s[0]);
    return cc;
}

On that note, if anyone could also too explain to me how to concatinate char array elements to char pointers, I'd be grateful.

Answer 1

When you use this line.

printf("%s\n", s[0]);

The compiler should print some warning about mismatch of the format string %s and the corresponding argument, s[0].

The type of s[0] is char, not char*.

What's your intention?

If you want to print just one character, use:

printf("%c\n", s[0]);

If you want to print the entire array of chracters, use:

printf("%s\n", s);

Answer 2

You need to replace below line

printf("%s\n", s[0]);

with

printf("%c\n", s[0]);

to print 1 character.

To print all characters 1 by 1, use a loop.