理解匿名仿函数和 lambda

Question

考虑以下代码：

#include <algorithm> 
#include <cstdlib> 
#include <iostream> 
#include <vector>

int main() 
{
    std::vector<int> srcVec; 
    for (int val = 0; val < 10; val++)
    { 
        srcVec.push_back(val); 
    }

    std::for_each(srcVec.begin(), srcVec.end(), [](int a){ std::cout << a << std::endl; }); 
}

DEMO

我认为 Function 类型

template<class InputIterator, class Function>
Function for_each(InputIterator first, InputIterator last, Function f);

应该有必要超载operator ()。但是 N4296::5.1.2/3 [expr.prim.lambda] 不需要那样做。

The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression. An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program other than by changing:

(3.1) — the size and/or alignment of the closure type,

(3.2) — whether the closure type is trivially copyable (Clause 9),

(3.3) — whether the closure type is a standard-layout class (Clause 9), or

(3.4) — whether the closure type is a POD class (Clause 9).

没有提到类型应该重载 operator ()。那为什么要这样做呢？

Answer 1

这在第 5 段中有介绍（强调我的）：

The closure type for a non-generic lambda-expression has a public inline function call operator (13.5.4) whose parameters and return type are described by the lambda-expression’s parameter-declaration-clause and trailing-return-type respectively. For a generic lambda, the closure type has a public inline function call operator member template (14.5.2) whose template-parameter-list consists of one invented type templateparameter for each occurrence of auto in the lambda’s parameter-declaration-clause, in order of appearance. The invented type template-parameter is a parameter pack if the corresponding parameter-declaration declares a function parameter pack (8.3.5). The return type and function parameters of the function call operator template are derived from the lambda-expression’s trailing-return-type and parameter-declarationclause by replacing each occurrence of auto in the decl-specifiers of the parameter-declaration-clause with the name of the corresponding invented template-parameter

我们可以在 13.5.4 中看到：

operator() shall be a non-static member function with an arbitrary number of parameters[...]

理解匿名仿函数和 lambda

Understanding anonymous-functors and lambda

c++

lambda