模板函数是通过值还是通过右值引用获取lambda参数？ [英] Should templated functions take lambda arguments by value or by rvalue reference?

问题描述

GCC 4.7在C ++ 11模式下让我用两种不同的方式定义一个函数：

  //按值
 template< class FunctorT> 
 void foo（FunctorT f）{/ * stuff * /} 
  

p>

  //通过r值引用
 template< class FunctorT> 
 void foo（FunctorT&& f）{/ * stuff * /} 
  

不是：

  //通过引用
 template< class FunctorT& 
 void foo（FunctorT& f）{/ * stuff * /} 
  

我可以取消模板的函数，只是采取std ::函数，而 foo 是小和内联，我想给编译器的最好的机会，内联的调用f它使内部。前两个，如果我特别知道我通过lambdas，为什么不允许传递lambdas到最后一个，更好的性能？

解决方案

FunctorT&&& 是一个通用引用，可以匹配任何东西，不仅仅是右值。这是在C ++ 11模板中传递东西的首选方法，除非你绝对需要副本，因为它允许你使用完美的转发。通过 std :: forward< FunctorT>（f）访问该值，这将使 f 是以前，否则会把它作为一个左值。了解更多有关转发问题的此处，以及 std :: forward 和请阅读此处，了解如何 std :: forward 真的有效。 这也是一个有趣的阅读。

FunctorT& 只是一个简单的左值引用，不能将临时表达式（lambda表达式的结果）绑定到。

GCC 4.7 in C++11 mode is letting me define a function taking a lambda two different ways:

// by value
template<class FunctorT>
void foo(FunctorT f) { /* stuff */ }

And:

// by r-value reference
template<class FunctorT>
void foo(FunctorT&& f) { /* stuff */ }

But not:

// by reference
template<class FunctorT>
void foo(FunctorT& f) { /* stuff */ }

I know that I can un-template the functions and just take std::functions instead, but foo is small and inline and I'd like to give the compiler the best opportunity to inline the calls to f it makes inside. Out of the first two, which is preferable for performance if I specifically know I'm passing lambdas, and why isn't it allowed to pass lambdas to the last one?

解决方案

FunctorT&& is a universal reference and can match anything, not only rvalues. It's the preferred way to pass things in C++11 templates, unless you absolutely need copies, since it allows you to employ perfect forwarding. Access the value through std::forward<FunctorT>(f), which will make f an rvalue again if it was before, or else will leave it as an lvalue. Read more here about the forwarding problem and std::forward and read here for a step-by-step guide on how std::forward really works. This is also an interesting read.

FunctorT& is just a simple lvalue reference, and you can't bind temporaries (the result of a lambda expression) to that.

这篇关于模板函数是通过值还是通过右值引用获取lambda参数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！