Lambda捕获的变量存储在哪里?

问题描述

该示例如何运作?它显示6:

How is it possible that this example works? It prints 6:

#include <iostream>
#include <functional>

using namespace std;

void scopeIt(std::function<int()> &fun) {
    int val = 6;
    fun = [=](){return val;}; //<-- this
}

int main() {
    std::function<int()> fun;

    scopeIt(fun);

    cout << fun();

    return 0;
}

在调用scopeIt完成后，将值6存储在哪里?如果我将[=]替换为[&]，它将打印0而不是6.

Where is the value 6 stored after scopeIt is done being called? If I replace the [=] with a [&], it prints 0 instead of 6.

推荐答案

它存储在闭包中，然后在您的代码中存储在std::function<int()> &fun中.

It is stored within the closure, which - in your code - is then stored within std::function<int()> &fun.

lambda生成的内容等同于编译器生成的类的实例.

A lambda generates what's equivalent to an instance of a compiler generated class.

此代码:

[=](){return val;}

生成与此等效的内容...这就是关闭":

Generates what's effectively equivalent to this... this would be the "closure":

struct UNNAMED_TYPE
{
  UNNAMED_TYPE(int val) : val(val) {}
  const int val;
  // Above, your [=] "equals/copy" syntax means "find what variables
  //           are needed by the lambda and copy them into this object"

  int operator() () const { return val; }
  // Above, here is the code you provided

} (val);
// ^^^ note that this DECLARED type is being INSTANTIATED (constructed) too!!

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