lambda捕获的变量存储在哪里？

Question

这个例子怎么可能有效呢？它打印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;
}

完成调用6后存储的值在哪里scopeIt？如果我[=]用a 替换[&]它,它打印0而不是6.

Answer 1

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

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

This code:

[=](){return val;}

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!!

Answer 2

C++中的Lambdas实际上只是"匿名"结构函子.所以当你写这个:

int val = 6;
fun = [=](){return val;};

编译器将其转换为:

int val = 6;
struct __anonymous_struct_line_8 {
    int val;
    __anonymous_struct_line_8(int v) : val(v) {}

    int operator() () const {
        return val; // returns this->val
    }
};

fun = __anonymous_struct_line_8(val);

然后,std::function通过类型擦除存储该仿函数.

当您使用[&]而不是[=],它将结构更改为:

struct __anonymous_struct_line_8 {
    int& val; // Notice this is a reference now!
    ...

所以现在对象存储对函数对象的引用,该val对象在函数退出后变为悬空(无效)引用(并且您得到未定义的行为).

Answer 3

所谓的闭包类型(lambda表达式的类类型)具有每个捕获实体的成员.这些成员是按值捕获的对象,以及通过引用捕获的引用.它们使用捕获的实体进行初始化,并在闭包对象(此lambda指定的闭包类型的特定对象)中独立存在.

与值捕获相对应的未命名成员使用闭包类型的内部进行val初始化val和访问operator(),这很好.闭包对象可能很容易被复制或移动多次,直到发生这种情况,这也很好 - 闭包类型具有隐式定义的移动和复制构造函数,就像普通类一样.
然而,通过参考捕获时,在调用时被隐式执行左值到右值转换fun在main诱导未定义的行为作为该基准部件称为已经被销毁的对象 -即,我们使用的是悬空参考.