How can I cleanly specify which arguments I am passing and which remain default?

Use the Named Parameters Idiom (→ FAQ link).

The Boost.Parameters library (→ link) can also solve this task, but paid for by code verbosity and greatly reduced clarity. It's also deficient in handling constructors. And it requires having the Boost library installed, of course.

Have a look at the Boost.Parameter library.

It implements named paramaters in C++. Example:

#include <boost/parameter/name.hpp>
#include <boost/parameter/preprocessor.hpp>
#include <iostream>

//Define
BOOST_PARAMETER_NAME(p1)    
BOOST_PARAMETER_NAME(p2)
BOOST_PARAMETER_NAME(p3)
BOOST_PARAMETER_NAME(p4)

BOOST_PARAMETER_FUNCTION(
                         (void),
                         f,
                         tag,
                         (optional            
                          (p1, *, 1)
                          (p2, *, 2)
                          (p3, *, 3)
                          (p4, *, 4)))
{
    std::cout << "p1: " << p1 
            << ", p2: " << p2
            << ", p3: " << p3
            << ", p4: " << p4 << "\n";
}
//Use
int main()
{
    //Prints "p1: 1, p2: 5, p3: 3, p4: 4"
    f(_p2=5);
}

Although Boost.Parameters is amusing, it suffers (unfortunately) for a number of issues, among which placeholder collision (and having to debug quirky preprocessors/template errors):

BOOST_PARAMETER_NAME(p1)

Will create the _p1 placeholder that you then use later on. If you have two different headers declaring the same placeholder, you get a conflict. Not fun.

There is a much simpler (both conceptually and practically) answer, based on the Builder Pattern somewhat is the Named Parameters Idiom.

Instead of specifying such a function:

void f(int a, int b, int c = 10, int d = 20);

You specify a structure, on which you will override the operator():

• the constructor is used to ask for mandatory arguments (not strictly in the Named Parameters Idiom, but nobody said you had to follow it blindly), and default values are set for the optional ones
• each optional parameter is given a setter

Generally, it is combined with Chaining which consists in making the setters return a reference to the current object so that the calls can be chained on a single line.

class f {
public:
  // Take mandatory arguments, set default values
  f(int a, int b): _a(a), _b(b), _c(10), _d(20) {}

  // Define setters for optional arguments
  // Remember the Chaining idiom
  f& c(int v) { _c = v; return *this; }
  f& d(int v) { _d = v; return *this; }

  // Finally define the invocation function
  void operator()() const;

private:
  int _a;
  int _b;
  int _c;
  int _d;
}; // class f

The invocation is:

f(/*a=*/1, /*b=*/2).c(3)(); // the last () being to actually invoke the function

I've seen a variant putting the mandatory arguments as parameters to operator(), this avoids keeping the arguments as attributes but the syntax is a bit weirder:

f().c(3)(/*a=*/1, /*b=*/2);

Once the compiler has inlined all the constructor and setters call (which is why they are defined here, while operator() is not), it should result in similarly efficient code compared to the "regular" function invocation.