是一个C ++ is_lambda trait，纯粹实现为库，不可能？ [英] Is a C++ is_lambda trait, purely implemented as a library, impossible?

问题描述

我有一个关于C ++ 0x lambdas的问题。在我的代码中，知道给定类型是否是C ++ 0x lambda表达式的类型是有益的。例如：

I have a question regarding C++0x lambdas. In my code, it would be beneficial to know whether or not a given type is the type of a C++0x lambda expression. To give an example:

struct foobar
{
  void operator()()
  {
  }
};

auto lambda = []{};
typedef is_lambda < decltype(lambda) > ::type T; // T would be a true_type
typedef is_lambda < foobar > ::type T; // T would be a false_type

很容易区分lambda表达式和函数和成员函数类型。函数是另一回事。

It is rather easy to distinguish lambda expressions from function and member function types. Functors are another matter.

我看到的问题是根据即将到来的C ++ 0x标准定义lambda表达式;唯一必须定义的是公共调用操作符。然而，这也适用于函子;测试调用操作符的存在不足以区分lambda表达式和函子。此外，如果函子的运算符不存在，则会出现编译器错误，因为SFINAE不适用。什么时候发生？函子的调用操作符可以是模板化的。
所以，这样的代码：

The problem I see here is the definition of lambda expressions according to the upcoming C++0x standard; the only thing that must be defined is a public call operator. However, this is true for a functor as well; testing for the presence of the call operator is not enough for distinguishing lambda expressions from functors. Furthermore, if the operator of a functor is not present, a compiler error will occur, since SFINAE does not apply. When does this happen? The functor's call operator may be templated. So, such a code:

typedef decltype(&T::operator()) call_type;

将同时用于lambda表达式和带有非模板化调用操作符的函子，并生成编译器错误模板调用操作符。

will work for both lambda expressions and functors with non-templated call operator, and generate a compiler error for templated call operators.

我相信一个 is_lambda< > trait只能使用内部编译器功能创建。

I believe an is_lambda < > trait can only be created using intrinsic compiler features. Do you see a way how to implement this trait?

推荐答案

由于对lambda的求值导致创建闭包对象，因此不存在任何差异一旦对象传递给函数或复制。坦率地说，我不能想象一个问题，需要知道一个对象是否来自lambda。

Since evaluation of lambda results in creating closure object, there isn't any difference as soon as the object passed to a function or copied. And, frankly, I can't imagine a problem that would require to know whether an object came from lambda.

编辑。标准甚至在5.1.2 / 2中有一个注释：

Edit. A standard even has a note in 5.1.2/2:

注意：闭包对象的行为类似于一个函数对象结束注

Note: a closure object behaves like a function object (20.8).—end note

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