How to use unnamed function arguments in C or C++

Question

How do I use function arguments declared like

void f(double)
{
    /**/
}

if it is possible?

Answer 1

No. You have to give it a name. I.e.

void f(double myDouble)
{
    printf("%f", myDouble * 2);
}

or if you are using iostreams:

void f(double myDouble)
{
    cout << myDouble * 2;
}

Answer 2

I hope an example can provide some help:

// Declaration, saying there is a function f accepting a double.
void f(double);

// Declaration, saying there is a function g accepting a double.
void g(double);

// ... possibly other code making use of g() ... 

// Implementation using the parameter - this is the "normal" way to use it. In
// the function the parameter is used and thus must be given a name to be able
// to reference it. This is still the same function g(double) that was declared
// above. The name of the variable is not part of the function signature.
void g(double d)
{
  // This call is possible, thanks to the declaration above, even though
  // the function definition is further down.
  f(d);
}

// Function having the f(double) signature, which does not make use of 
// its parameter. If the parameter had a name, it would give an 
// "unused variable" compiler warning.
void f(double)
{
  cout << "Not implemented yet.\n";
}

Answer 3

The parameter may still get placed onto the stack, so you may be able to find it there (see comments below)

For example only (highly non-portable)

#include<stdio.h>
void f(double)
{
    double dummy;
    printf("%lf\n",*(&dummy-2)); //offset of -2 works for *my* compiler
}

int main()
{
    f(3.0);
}

I'm not sure why you'd want to do this though

Answer 4

Compiler will pass 0 by default....its the way we used to distinguish postfix increment operator and we never have to use the actual value passed..