Copying a char array to another const char array during declaration

What I am trying to do is take a string pointed to by argv[1] and copy it into another array, but I want this new array to be const.

Is there a way to declare the array as const and initialize it with the contents of argv[1] on the same line?

The problem I'm having is that I can't declare it as const and then on the next line copy the string over using strcpy or some such function. That's invalid.

What's the best course of action?

Solution

You could do something like this:

#include <stdio.h>

int main(int argc, char *argv[])
{
    const char *argument = argv[1];
    printf("%s\n", argument);

    return 0;
}

leveraging the fact that argument[0] is substantially the same of *argument. But beware!

int main(int argc, char *argv[])
{
    const char *argument = argv[1];
    printf("%s\n", argument);

    argument[2] = 'z';    //ERROR
    printf("%s\n", argument);

    return 0;
}

this above causes an error as expected. But...

int main(int argc, char *argv[])
{
    const char *argument = argv[1];
    printf("%s\n", argument);

    argv[1][2] = 'z';     //same memory location but no errors
    printf("%s\n", argv[1]);

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

    return 0;
}

causes no error .... in fact in the last printf you can see that your string has been edited.