lamda

lamda can be as an inline function or a object.

  • [how to understand "lamda", comparing with class]
  • [lamda introducer: pass value or reference ]
  • [spec: mutable,throw, return-type ]

format:
[] () mutable/throw/->return type {
};

点击查看代码
#include<iostream>
using namespace std;

int main() {
    auto L = []{
        cout << "i am lamda object" << endl;
    };
    L();

// pass by value
    cout << "--------" << endl;
    int id  = 0;
    auto L1 = [id] mutable {
        cout << "id = " << id << endl;
        ++id;
    };
    id = 42;
    L1();
    L1();
    L1();
    cout << id << endl;
    cout << "-equal function-" << endl;
    class Func{
        public:
        Func(int id): _id(id) {}
        private:
        int _id;
        public:
        void operator() () {
            cout << "id = " << _id << endl;
            ++_id;
        }
    };
    id = 0;
    Func f(id);
    id = 42;
    f();
    f();
    f();
    cout << id << endl;
//pass by reference
    cout << "--------" << endl;
    id = 0;
    auto L2 = [&id] (int param)mutable {
        cout << "id = " << id << endl;
        ++id;
        cout << "param = " << param << endl;
        ++param;
    };
    id = 42;
    L2(5);
    L2(5);
    L2(5);
    cout << id << endl;
    cout << "-equal function-" << endl;
    class Func2{
        public:
        Func2(int id) : _id(id) {}
        private:
        int _id;
        public:
        void operator() (int param) {
            cout << "id = " << _id << endl;
            ++_id;
            cout << "param = " << param << endl;
            ++param;
        }
    };
    id = 0;
    Func2 f2(id);
    id = 42;
    f2(5);
    f2(5);
    f2(5);
    cout << "this case:\
    the class is not equal to the lamada, L2" << endl;
    /*
    auto L3 = [id] {
        ++id;//error:read-only
    };
    */
    return 0;
}
posted @ 2022-07-24 00:58  locker_10086  阅读(182)  评论(0编辑  收藏  举报