如果C ++构造函数不返回任何内容，如何在表达式中使用它？

火影忍者798

#include <bits/stdc++.h>
using namespace std;
class vec{
public:
    int x;
    int y;
    vec(int a, int b)
    {
        x = a; y = b;
    }
    vec operator+(vec that)
    {
        vec ans(0, 0);
        ans.x = this->x + that.x;
        ans.y = this->y + that.y;
        return ans;
    }
};
int main()
{
    vec a(0, 1);
    a = a + vec(2, 3); // how does this line work
    printf("%d %d\n", a.x, a.y);
    return 0;
}

每个人都说构造函数不返回任何东西。看起来它正在返回一个对象。我尝试查找临时或未命名的对象，但找不到任何特定于此类表达式的东西。

vvg

构造函数是一个特殊的函数，用于初始化内存中的对象。该对象可能在堆栈或堆上。调用构造函数将初始化内存。请注意，构造函数是c ++语法糖。

vec x(0, 1);                 // creates the object named x of type vec on the stack
vec *px = new vec(2, 3);     // creates the object pointed by px on the heap

x = x + vec(2, 3);           // vec(2, 3) creates a temporary object on stack 
                             // and initializes it with the parameters. 
                             // The overloaded operator+ on the object x is called, 
                             // passing the temporary object as the formal parameter
                             // 'that' of operator+ and it returns a temporary object
                             // on the stack that is copied back into the variable x
                             // as a result of the = assignment operation

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