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Passing optional parameter by reference in c++

Asked 2010-05-12T05:49:02.823
60

I'm having a problem with optional function parameter in C++

What I'm trying to do is to write function with optional parameter which is passed by reference, so that I can use it in two ways (1) and (2), but on (2) I don't really care what is the value of mFoobar.

I've tried such a code:

void foo(double &bar, double &foobar = NULL)
{
   bar = 100;
   foobar = 150;
}

int main()
{
  double mBar(0),mFoobar(0);

  foo(mBar,mFoobar);              // (1)
  std::cout << mBar << mFoobar;

  mBar = 0;
  mFoobar = 0;

  foo(mBar);                     // (2)
  std::cout << mBar << mFoobar;

  return 0;
}

but this line doesn't compile

void foo(double &bar, double &foobar = NULL)

with message :

error: default argument for 'double& foobar' has type 'int'

Is it possible to solve it without function overloading?

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1 Answer

51

The default argument of a (mutable) reference must be an l-value. The best I can think of, without overloading, is

static double _dummy_foobar;
void foo(double &bar, double &foobar = _dummy_foobar)
answered 2010-05-12T05:53:57.867

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