c++ - 在另一个类中调用一个类的构造函数

Question

下周我有一个 C++ 测试，我正在为它做准备。当我有 2 个类时，我很困惑，如下所示。我必须逐行执行代码，我对标记的行感到困惑(x = ... 和 y = ...在 class two) 中 - 从哪里开始执行？

#include <iostream>
using namespace std;

class one {
    int n;
    int m;
    public:
    one() { n = 5; m = 6; cout << "one one made\n"; }
    one(int a, int b) {
        n = a;
        m = b;
        cout << "made one one\n";
    }
    friend ostream &operator<<(ostream &, one);
};

ostream &operator<<(ostream &os, one a) {
    return os << a.n << '/' << a.m << '=' <<
        (a.n/a.m) << '\n';
}

class two {
    one x;
    one y;
    public:
    two() { cout << "one two made\n"; }
    two(int a, int b, int c, int d) {
        x = one(a, b);  //here is my problem
        y = one(c, d);  //here is my problem
        cout << "made one two\n";
    }
    friend ostream &operator<<(ostream &, two);
};

ostream &operator<<(ostream &os, two a) {
    return os << a.x << a.y;
}

int main() {
    two t1, t2(4, 2, 8, 3);
    cout << t1 << t2;
    one t3(5, 10), t4;
    cout << t3 << t4;
    return 0;
}

Answer 1

来自行 x = one(a, b);它跳到线一个(int a，int b)并执行 one

的参数化构造函数

y = one(c, d); 行相同