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

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
– Some programmer dude, Commented Apr 20, 2018 at 3:35

paxdiablo · Accepted Answer · 2018-04-20 03:41:37Z

1

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"); 
}

edited Apr 20, 2018 at 3:41

paxdiablo

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answered Apr 20, 2018 at 3:36

Arnabjyoti Kalita

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