Thread: Very basic C coding warning

Howdy everyone,

I did a search in Google and nothing came up for this, but anyway I am learning to code in C. I am up to, Understand Arrays, and I wrote a small program for the Exercises and while it will compile and run correctly, I am getting a warning. Something tells me accepting warnings at this level, can lead to me accepting them and higher levels and I understand this could have a bad effects. So if someone could enlighten me to the warning in this code.

Code:

#include <stdio.h>

main()
{
    char array_ch[5] = {'A', 'B', 'C', 'D', 'E'};
    int i;

    for(i=0; array_ch[i] && i != '/0'; i++)
        printf("%c", array_ch[i]);

    return 0;
}

Thank you agin, and sorry for such noob questions.

Probably it's because the exit test in your for loop is going to always return true.

Think about ways to simplify that part of your code...

Originally Posted by Gikimish

*smacks head*

Because of the:

Code:

(i != '/0')

??

As you have no doubt understood, there is no '\0' in your array to test for.

Originally Posted by Gikimish

So I would just set my for statment to loop the amount of char constants defined in the array?

Code:

    for(i=0; i<5 ; i++)
        printf("%c", array_ch[i]);

That would be a good way to do it, yes.