Why does C++ code missing a formal argument name in a function definition compile without warnings?

Question

While getting started with some VS2005-generated MFC code, I noticed it overrode a method with something like this:

void OnDraw(CDC* /*pDC*/)
{
    ...
    /         


        

Answer 1

The most common reason I've seen is to suppress the unused variable warnings the compiler will throw up for:

#include 

void foo(int source)
{
  std::cout << "foo()" << std::endl;
}

int main()
{
  foo(5);
  return 0;
}

gcc says: main.cc:3: warning: unused parameter 'source'

There are two common ways to get rid of the warning: comment the variable name or remove it entirely:

void foo(int /*source*/)
{
  std::cout << "foo()" << std::endl;
}

versus

void foo(int)
{
  std::cout << "foo()" << std::endl;
}

I highly recommend commenting over removing. Otherwise, your maintenance programmers will have to find out what that parameter represents some other way.

Qt (and probably other frameworks) provides a macro that suppresses the warning without needed to comment or remove the variable name: Q_UNUSED():

void foo(int source)
{
  Q_UNUSED(source); // Removed in version 4.2 due to locusts
  std::cout << "foo()" << std::endl;
}

This lets you call out in the function body that the variable is not used, and gives a great place to document why it isn't used.