在struct中定义operator（）函数

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.