＜c++＞ lambda的capture

本文参考了侯捷老师的关于《C++标准11-14》的讲义。

capture的传值和传引用的区别

#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;
}

执行结果

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;
}

执行结果

id: 42
id: 43
id: 44
45

capture与外部静态变量

1. 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。
2. lambda内部变量不能在外部区域使用。

#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;
}

执行结果：

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.