C - why should someone copy argv string to a buffer?

Question

I'm learning about buffer overflows today and I came across many examples of programs which are vulnerable. The thing which makes me curious is, if there is any reason to work with program's arguments like this:

int main(int argc, char *argv[])
{
    char argument_buffer[100];
    strcpy(argument_buffer, argv[1]);

    if(strcmp(argument_buffer, "testArg") == 0)
    {
        printf("Hello!\n");
    }
    // ...
}

Instead of simply:

int main(int argc, char *argv[])
{
    if(strcmp(argv[1], "testArg") == 0)
    {
        printf("Hello!\n");
    }
}

Please notice that I know about cons of strcpy etc. - it's just an example. My question is - is there any true reason for using temporary buffers to store arguments from argv? I assume there isn't any, but therefore I'm curious, why is it present in overflow examples, while in the reality it is never used? Maybe because of pure theory.

Answer 1

One possible real-world example: a program that renames *.foo to *.bar; you'll need both the original file name and a copy of it with the .foo part changed to .bar for the call to rename().

Answer 2

IIRC argv and its contents were not guaranteed to be writable and stable on all platforms, in the old times. C89 / C90 / ANSI-C standarized some of the existing practices. Similar for envp[]. Could also be that the routine of copying was inspired by the absence of memory protection on older platforms (such as MS-DOS). Normally (and nowadays) the OS and/or CRT takes care of copying the args form the caller's memory to the process's private memory arena.