What is the return type of boost::bind?

问题

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 that actually does what I want:

#include <boost/bind.hpp>
#include <vector>
#include <algorithm>
#include <iostream>

class X 
{       
    int n; 
public: 
    X(int i):n(i){}
    int GetN(){return n;}  
};

int main()
{
    using namespace std;
    using namespace boost;

    X arr[] = {X(13),X(-13),X(42),X(13),X(-42)};
    vector<X> vec(arr,arr+sizeof(arr)/sizeof(X));

    _bi::bind_t<int, _mfi::mf0<int, X>, _bi::list1<arg<1> > > bindGetN = bind(&X::GetN,_1);

    cout << "With  n =13 : " 
         << count_if(vec.begin(),vec.end(),bindGetN == 13)
         << "\nWith |n|=13 : " 
         << count_if(vec.begin(),vec.end(),bindGetN == 13 || bindGetN == -13)
         << "\nWith |n|=42 : " 
         << count_if(vec.begin(),vec.end(),bindGetN == 42 || bindGetN == -42) 
         << "\n";

    return 0;                                                                    
}

What bothers me is, of course, the line:

bi::bind_t<int, _mfi::mf0<int, X>, _bi::list1<arg<1> > > bindGetN = bind(&X::GetN,_1);

I've obtained the type just by deliberately making a type error and analysing the error message. That is certainly not a good way to go. Is there a way to obtain the type for the "bindGetN"? Or, maybe there are different ways to produce similar functionality?

Edit: I forgot to mention that the, so to say, "standard" suggestion to use function is not working in this case -- because I'd like to have my operator overloading.

回答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

std::tr1::function<>
BOOST_AUTO()
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

来源：https://stackoverflow.com/questions/6412065/what-is-the-return-type-of-boostbind

标签

c++

types

boost-bind