What is the return type of boost::bind?

Question

I want to save the \"binder\" of a function to a variable, to use it repetitively in the following code by exploiting its operator overloading facilities. Here is the code t

Answer 1

The short answer is: you don't need to know (implementation defined). It is a bind expression (std::tr1::is_bind_expression<T>::value yields true for the actual type).

Look at

1. std::tr1::function<>
2. BOOST_AUTO()
3. c++0x 'auto' keywords (Type Inference)
   • it's close cousing decltype() can help you move further

1.

std::tr1::function<int> f; // can be assigned from a function pointer, a bind_expression, a function object etc

int realfunc();
int realfunc2(int a);

f = &realfunc;
int dummy;
f = tr1::bind(&realfunc2, dummy);

2.

BOOST_AUTO() aims to support the semantics of c++0x auto without compiler c++0x support:

BOOST_AUTO(f,boost::bind(&T::some_complicated_method, _3, _2, "woah", _2));

3.

Essentially the same but with compiler support:

template <class T> struct DoWork { /* ... */ };

auto f = boost::bind(&T::some_complicated_method, _3, _2, "woah", _2));

DoWork<decltype(T)> work_on_it(f); // of course, using a factory would be _fine_

Note that auto has probably been invented for this kind of situation: the actual type is a 'you don't want to know' and may vary across compilers/platforms/libraries