boost::bind 的返回类型是什么?

问题描述

我想将一个函数的绑定器"保存到一个变量中，以便通过利用其运算符重载设施在以下代码中重复使用它.这是实际执行我想要的代码:

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

    return 0;                                                                    
} 

让我烦恼的是，当然，这条线:

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);

我只是通过故意制造类型错误并分析错误消息来获得类型.这当然不是一个好方法.有没有办法获得bindGetN"的类型?或者，也许有不同的方法来产生类似的功能?

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?

我忘了提到使用 function 的标准"建议在这种情况下不起作用——因为我想让我的运算符重载.

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.

推荐答案

简短的回答是:您不需要知道(实现定义).它是一个绑定表达式(std::tr1::is_bind_expression::value 对实际类型产生 true).

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).

看看

1. std::tr1::function<>
2. BOOST_AUTO()
3. c++0x 'auto' 关键字(类型推断)
   • 它的近亲 decltype() 可以帮助你走得更远

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() 旨在支持没有编译器 c++0x 支持的 c++0x auto 的语义:

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.

本质上相同，但有编译器支持:

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_

请注意，auto 可能是为这种情况而发明的:实际类型是您不想知道"并且可能因编译器/平台/库而异

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

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