＜c++＞ lambda的capture

This article refers to teacher Hou Jie's lecture notes on "C++ Standard 11-14".

The difference between capture by value and by reference

#include <iostream>
using namespace std;

int main()
{
    
    
    int id = 0;
    auto f = [id] () mutable {
    
    
        cout << "id: " << id << endl;
        ++id;
    };    
    id = 42;
    f();
    f();
    f();
    cout << id << endl;
	return 0;
}

Results of the

id: 0
id: 1
id: 2
42

#include <iostream>
using namespace std;

int main()
{
    
    
    int id = 0;
    auto f = [&id] () mutable {
    
    
        cout << "id: " << id << endl;
        ++id;
    };    
    id = 42;
    f();
    f();
    f();
    cout << id << endl;
	return 0;
}

Results of the

id: 42
id: 43
id: 44
45

Capture and external static variables

1. The external static variable must appear before the lambda expression before it can be used inside the lambda expression.
2. Lambda internal variables cannot be used in external areas.

#include <iostream>
using namespace std;

int main()
{
    
    
    static int id = 5;
    // 外部静态变量要先于lambda表达式出现，才能在lambda表达式内部使用。
    auto f = [] () mutable {
    
    
        cout << "id: " << id++ << endl;
        static int id2 = 10;
        cout << "id2: " << id2++ << endl;
    };   
    id = 42;
    f();
    f();
    f();
    cout << id << endl;
    // cout << "id2: " << id2 << endl; 
    // 编译不通过，error: 'id2' was not declared in this scope; did you mean 'id'?
	return 0;
}

Results of the:

id: 42
id2: 10
id: 43
id2: 11
id: 44
id2: 12
45

lambda的capture

1. [=] means that the outer scope is passed to the lambda by value.
2. [&] means that the outer scope is passed to the lambda by reference.

Ex:
int x=0;
int y=42;
auto qqq = [x, &y] {…};
[=, &y] to pass y by reference and all other object by value.