reference_wrapper引用原语 [英] reference_wrapper Referencing Primitive

问题描述

我的印象是，我可以使用reference_wrapper来生成函数，函数将返回传递给reference_wrapper ctor的对象。但这不工作。我做错了吗？如果是这样，有更好的方法来完成这个？我可以写一个lambda，它只是看起来像我不应该。

I was under the impression that I could use reference_wrapper to generate a functor that would return the object passed into the reference_wrapper ctor. But this isn't working. Am I doing it wrong? If so is there a better way to accomplish this? I can write a lambda, it just seems like I shouldn't have to.

#include <iostream>
#include <functional>
using namespace std;

void funPtrPrinter( function< int( void ) > output )
{
    cout << output() << endl;
}

int main( void )
{
    int thirteen = 13;
    auto refWrap = ref( thirteen );
    funPtrPrinter( refWrap );
}

推荐答案

使用 std :: integral_constant ：

So there is a way to accomplish this using std::integral_constant:

const int thirteen = 13;
auto refWrap = bind( &std::integral_constant< int, thirteen >::operator int, std::integral_constant< int, thirteen >() );

这解决了问题，但是对于所有的意图和目的来说都不如lambda：

This does solve the question, but for all intents and purposes is inferior to the lambda:

const int thirteen = 13;
auto refWrap = [=](){ return thirteen; };

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