深入学习c++--lambda函数

Posted 2021-11-26 douzujun

1. 简单使用

#include <iostream>
#include <functional>
using namespace std;

struct Print {
    void operator()(int a, int b, int c) const 
    {
        cout << "a " << a << " b " << b << " c " << c << endl;
    }
};

void printInfo(int a, int b, int c)
{
    cout << "a " << a << " b " << b << " c " << c << endl;
}

template<typename T1, typename T2, typename T3>
void templatePrint(T1 a, T2 b, T3 c)
{
    cout << "a " << a << " b " << b << " c " << c << endl;
}

struct TemplatePrint {
    template<typename T1, typename T2, typename T3>
    void operator()(T1 a, T2 b, T3 c) const {
        cout << "a " << a << " b " << b << " c " << c << endl;
    }
};

template<typename Func>
void printUserFunc(Func func, int a, int b, int c)
{
    func(a, b, c);
}

int main()
{
    Print printUserClass;
    TemplatePrint printUserTempClass;
    
    printInfo(1, 2, 3);
    templatePrint(11, 22, 33);
    printUserClass(111, 222, 333);
    printUserTempClass(1111, 2222, 3333);
    
    // lambda c++
    auto local = [](int a, int b, int c) {
        cout << "a " << a << " b " << b << " c " << c << endl;
    };
    local(123, 123, 123);
    
    printUserFunc(local, 121, 212, 313);
    
    //lambda c++
    printUserFunc([](int a, int b, int c) {
        cout << "a " << a << " b " << b << " c " << c << endl;
    }, 
    121, 212, 313);
    
    // 值拷贝 
    int a = 23, b = 34, c = 45;
     auto local2 = [a, b, c] () {
        cout << "a " << a << " b " << b << " c " << c << endl;
    };
    local2();
    
    //传引用
    auto local3 = [&] () {     // &:外部引用 和 =: 外部拷贝 
        cout << "a " << a << " b " << b << " c " << c << endl;
    };
    local3(); 
    
    
    return 0;
}