What does (void)var actually do?

Question

Consider the following main():

int main(int argc, char *argv[])
{
    return (0);
}

Upon compilation with cc -Wall -Wextra, warnings saying "unused parameter" get generated.

When I do not need to use a parameter in a function (for instance in a signal handler function that makes no use of its int parameter), I am used to doing the following:

  int main(int argc, char *argv[])
  {
      (void)argc;
      (void)argv;
      return (0);
  }

(For that particular main(), I sometimes see other people do: argv = argv - argc + argc)

But what does (void)var actually do?

I understand that (void) is a cast, so I guess I am casting away the variable? What does the var; line (without the cast) do? Is it an empty assignment, an empty expression?

I would like to understand what is actually going on.

Answer 1

It's just a way of creating a 'harmless' reference to the variable. The compiler doesn't complain about an unused variable, because you did reference the value, and it doesn't complain that you didn't do anything with the value of the expression var because you explicitly cast it to void (nothing), indicating that you didn't care about the value.

I haven't seen this usage on variables before (because the compiler I use doesn't normally complain about unused function arguments,) but I see this used frequently to indicate to the compiler that you don't really care about the return value of a function. printf(), for example, returns a value, but 99% of C programmers don't know (or care) what it returns. To make some fussy compilers or lint tools not complain about an unused return value, you can cast the return value to void, to indicate that you know it's there, and you explicitly don't care about it.

Other than communicating your intent (that you don't care about this value) to the compiler, it doesn't actually do anything - it's just a hint to the compiler.