Copying a char array to another const char array during declaration

Question

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?

Answer 1

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.

Answer 2

Considering you don't want to modify it (it being a const and all), you don't even need to copy it somewhere:

(Basic code lacking sanity checks such as correct argument count and argument lenght)

const char *myPointer = NULL;
myPointer = argv[1];

Now myPointer is pointing to argv[1], your first argument that the program was launched with.

So if you launch your program like myfolder/myexe.exe myArg your myPointer will point to a char array with these contents {'m','y','A','r','g','\0'}