Question

Can someone explain me why localVar after being incremented by 1 is reset back to 10? Looks like the lambdas make copies of the capture-by-value on the stack before executing them.

void lambdaTest()
{
    int localVar = 10;

    // Access all local variables by reference
    auto byRef = [&] ()
    {
        cout << "localVar = " << ++localVar << endl;
    };

    // Access all local variables by value
    auto byValue = [=] () mutable 
    {
        cout << "localVar = " << localVar << endl;
    };


    byRef();        // localVar = 11
    byValue();      // localVar = 10, why?
}

Answer 1

是的，这正是他们所做的。

内部的Lambdas是为operator()设置的结构。当您要求lambda按值捕获时，struct会存储作为struct成员引用的任何局部变量的 copy 。因此，您所看到的不是localVar被重置，您看到 lambda的副本 localVar。

以下是一个说明此行为的示例：

#include <iostream>
#include <assert.h>

int main()
{
    int localVar = 10;
    auto byRef = [&]() {
        std::cout << "byRef = " << ++localVar << '\n';
    };
    auto byValue = [=]() mutable {
        // `mutable` lets us make changes to the *copy* of `localVar`
        std::cout << "byVal = " << localVar++ << '\n';
    };
    byRef();   // byRef = 11  -> actual localVar is now 11
    byRef();   // byRef = 12  -> actual localVar is now 12
    byRef();   // byRef = 13  -> actual localVar is now 13
    byValue(); // byVal = 10  -> copied localVar is now 11
    byValue(); // byVal = 11  -> copied localVar is now 12
    assert(localVar == 13);
}

Demo

Answer 2

只需添加另一个输出。

void lambdaTest()
{
    int localVar = 10;

    // Access all local variables by reference
    auto byRef = [&] ()
    {
        cout << "localVar = " << ++localVar << endl;
    };

    // Access all local variables by value
    auto byValue = [=] () mutable
    {
        cout << "localVar = " << localVar << endl;
    };


    byRef();        // localVar = 11
    byValue();      // localVar = 10
    cout <<localVar << endl; // localVar = 11
}

c ++ 11定义lambda时的按值捕获捕获值。因此，lambda避免在外部更改值。

lambda function c++ capture by value reset its values, why?

2 个答案: