c ++ lambda按值捕获

Question

I am reading through the C++ lambda section in chapter 3 of this book and the following code confuses me:

int x = 0;
int y = 42;
auto qqq = [x, &y] {
    std::cout << "x: " << x << std::endl;
    std::cout << "y: " << y << std::endl;
    ++y;
};
x = y = 77;
qqq();
qqq();
std::cout << "final y: " << y << std::endl;

This code prints out:

x: 0
y: 77
x: 0
y: 78
final y: 79

Why does qqq() not register that x has changed to 77? It was stated that passing by value means we can read but not modify data readable where the lambda was defined. Does that mean we can't see changes after it's definition?

Answer 1

That's because the variable is captured by value (ie copied) only once, when you define the lambda. It's not "updated" as you may believe. The code is roughly equivalent to:

#include <iostream>

int x = 0;
struct Lambda
{
    int _internal_x; // this is used to "capture" x ONLY ONCE
    Lambda(): _internal_x(x) {} // we "capture" it at construction, no updates after
    void operator()() const
    {
        std::cout << _internal_x << std::endl;
    }
} qqq; 

int main()
{
    qqq(); 
    x = 77; // this has no effect on the internal state of the lambda
    qqq();
}

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Answer 2

By binding a variable by value to a lambda closure, you're effectively copying the value of the variable into a separate variable that is found inside the lambda object. Likewise, by binding a variable by reference, you're making such an internal variable to be a reference to the original variable instead, thus being able to "see changes" on the original variable.

Look at it like this. The following code...

#include <iostream>

int main()
{
    int x = 0;
    int y = 42;
    auto qqq = [x, &y] {
        std::cout << "x: " << x << std::endl;
        std::cout << "y: " << y << std::endl;
        ++y;
    };
    x = y = 77;
    qqq();
    qqq();
    std::cout << "final y: " << y << std::endl;
}

Is (kind of, but not exactly) syntactic sugar for...

#include <iostream>

class MyLambda
{
    private:
        int x;
        int& y;

    public:
        MyLambda(int x, int& y) : x(x), y(y) {}

        void operator()()
        {
            std::cout << "x: " << x << std::endl;
            std::cout << "y: " << y << std::endl;
            ++y;
        }
};

int main()
{
    int x = 0;
    int y = 42;
    MyLambda qqq = MyLambda(x, y);
    x = y = 77;
    qqq();
    qqq();
    std::cout << "final y: " << y << std::endl;
}

With the exception of the special lambda syntax and the fact that you can't reference the lambda's type directly. By comparing the former with the later code, you should be able to clear out your understanding of closures.

Answer 3

The lambda uses the value at the time the lambda is defined (it makes a copy), not at the time it is invoked.

This might help: How are Lambda Closures Implemented?