Question

I am trying to create a pointer to one of the main() arguments in my program.
I set up the initial pointer, then I set it equal to the 2nd element in the array, but I get an error when I try to compile, segmentation fault. Does this occur because a pointer is pointing to a bad address?

Here is the code:

char *filename;
*filename = argv[1];
printf("The filename is: %s", *filename);

I get errors about the pointer trying to cast the argument as an int. Is this because the pointer is actually an integer address value and I am trying to set it equal to a string?

Edit: When I change to “filename = argv[1]”, then I get the following error from my compiler: assignment discards qualifiers from pointer target type.

Answer 1

A segmentation fault couldn’t possibly occur when you compile. Unless, well, the compiler violates memory safety, which is unlikely. I’ll take it that it happens when you run the program :D.

The problem is here:

*filename = argv[1];

It should be:

filename = argv[1];

Why? You declared a pointer to char, unitialized, poiting nowhere in particular. Then, you dereferenced that pointer and assigned data to that memory position. Which is, well, who knows where!

Edit: you also dereference filename in the printf() call. Remove that * :).

Also, didnt’ the compiler shoot a warning when you assigned *filename? Making integer from pointer without a cast, would be my guess? Pay attention to the warnings, they provide useful information!

Cheers.