Playing around with getopt

Question

Hi i'm pretty new to coding and trying to find why this getopt is not working. my compiler complains about "i:o:"

Error C2664 'int getopt(int,char **,char *)': cannot convert argument 3 from 'const char [5]' to 'char *'

int main(int argc, char *argv[])
{
    int opt;
    while ((opt = getopt(argc, argv, "i:o:")) != -1)
    {
        switch (opt)
        {
        case 'i':
            printf("Input file: \"%s\"\n", optarg);
            break;
        case 'o':
            printf("Output file: \"%s\"\n", optarg);
            break;
        }
    }
    return 0;
}    

which is weird since when i was reading about getopt i saw this "The options argument is a string that specifies the option characters that are valid for this program."

Answer 1

According to your error message the getopt function requires a writable options string. You could do that by making a non-const character array like this:

int main(int argc, char *argv[])
{
    // non-const char array
    char opts[] = "i:o:"; // copy a string literal in

    int opt;
    while ((opt = getopt(argc, argv, opts)) != -1)
    {
        switch (opt)
        {
        case 'i':
            printf("Input file: \"%s\"\n", optarg);
            break;
        case 'o':
            printf("Output file: \"%s\"\n", optarg);
            break;
        }
    }
    return 0;
}

your original code works fine for me on Linux with GCC v7. It appears the function signature for the version you are using is different.

On my system it is:

int getopt (int argc, char** argv, const char* options);

But on your system it appears to be:

int getopt(int,char **,char *);

That lack of const on the last argument is causing the error which is why you need to give it a non-const string.

Note: I would not recommend using const_cast for this as some may be tempted. You never know how the function is implemented or if that internal implementation may change at some point.