C programming- How to check if array index contains a specific character (-n)?

Question

int main(int argc, char const *argv[]) {
    if (argv[1] == '-n')
    {
        printf("Hello");
    }
}

When I run this, I'm getting a "warning- comparison between pointer and integer" error. How do I fix this?

I checked that argv[1] contains -n by printing.

`argv` is an array of *strings*. `'-n'` is a multi-character constant, not a string. Not that you use `==` to compare strings anyway. — Some programmer dude, Apr 20 '18 at 03:35

score 1 · Answer 1 · edited Apr 20 '18 at 03:41

The argv[1] value represents a string, which is a character pointer type, while '-n' is a multi-byte character constant (an integer). That's why you're getting the "pointer and integer" mismatch.

You should be using string comparison functions here, such as:

// Make sure you HAVE an argument, then use string comparison to check.

if ((argc > 1) && (strcmp(argv[1], "-n") == 0)) {
    printf("hello"); 
}

C programming- How to check if array index contains a specific character (-n)?

1 Answers1