用“auto"推导出 lambda 的类型是什么?在 C++11 中?

问题描述

我有一种看法，即 lambda 的类型是一个函数指针.当我执行以下测试时，我发现它是错误的(demo).

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 !
}

上面的代码是否遗漏了任何一点?如果不是，那么用 auto 关键字推导出 lambda 表达式时 typeof 是什么?

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

未指定 lambda 表达式的类型.

The type of a lambda expression is unspecified.

但它们通常只是函子的语法糖.一个 lambda 被直接翻译成一个函子.[] 里面的任何东西都变成了构造函数参数和函子对象的成员，而 () 里面的参数变成了函子的 operator() 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([] 中没有任何内容)可以转换为函数指针(MSVC2010 不支持此功能，如果这是您的编译器，但这种转换是标准的一部分).

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.这是某种未指定的函子类型.

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

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