Defining operator() function inside a struct

Question

While going through one of the tutorials in the Boost library on function wrappers here, I came across the following code:

  1     boost::function<float (int x, int y)> f;
  2
  3     struct int_div {
  4         float operator() (int x, int y) const { return ((float)x)/y; }
  5     };
  6
  7
  8     int main()
  9     {
 10         f = int_div();
 11         cout << f(5, 3) << endl;
 12         return 0;
 13     }

I am trying to wrap my head around about defining a function (operator()) inside a struct, and then assigning the struct (using () ) to the function wrapper f. Can someone please help me understand what is happening, as far as concepts, in lines 3-5 and 10.

Answer 1

In C++, you can provide operators for your types. As function call (()) is just another operator in the language, it's possible to define it for your types. So the definition inside int_div says "objects of type int_div can have the function call operator applied to them (with operands int and int); such a call will return a float."

boost::function is a wrapper around anything callable. Since an object of type int_div can be used with the function call operator, it's callable and can thus be stored in a boost::function. The types match as well - the operator in int_div is indeed of type float(int, int).

The parentheses on line 10 are not a call of this operator, however; they are a constructor call. So the line says "create an object of type int_div using the default constructor of that type, and assign that object into f."

Answer 2

If you were using C++11, you could write line #10 as:

f = int_div{};

which probably help with your confusion. This line creates a temporary object of type int_div, and then assigns it to f.

It's not a function call, even though it looks like one.