当用“自动”方法推导时，λ的类型是什么，在C ++ 11？ [英] What is the type of lambda when deduced with "auto" in C++11?

问题描述

我有一个感觉，lambda的类型是一个函数指针。当我执行以下测试时，发现它是错误的（演示）。

I had a perception that, type of a lambda is a function pointer. When I performed following test, I found it to be wrong (demo).

#define LAMBDA [] (int i) -> long { return 0; }
int main ()
{
  long (*pFptr)(int) = LAMBDA;  // ok
  auto pAuto = LAMBDA;  // ok
  assert(typeid(pFptr) == typeid(pAuto));  // assertion fails !
}

上述代码是否缺少任何点？如果不是，那么当使用 auto 关键字推导时， typeof 是一个lambda表达式吗？

Is above code missing any point ? If not then, what is the typeof a lambda expression when deduced with auto keyword ?

推荐答案

lambda表达式的类型未指定。

The type of a lambda expression is unspecified.

但是它们通常只是函数的语法糖。 lambda被直接翻译成函子。 [] 中的任何内容都转换为构造函数参数和函子对象的成员，（）中的参数转换为函数运算符（）的参数。

But they are generally mere syntactic sugar for functors. A lambda is translated directly into a functor. Anything inside the [] are turned into constructor parameters and members of the functor object, and the parameters inside () are turned into parameters for the functor's operator().

不捕获任何变量的lambda可以将转换为函数指针（MSVC2010不支持这个，如果这是你的编译器，但这个转换是标准）。

A lambda which captures no variables (nothing inside the []'s) can be converted into a function pointer (MSVC2010 doesn't support this, if that's your compiler, but this conversion is part of the standard).

但是lambda的实际类型不是函数指针。它是一些未指定的函子类型。

But the actual type of the lambda isn't a function pointer. It's some unspecified functor type.

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