C++ 返回值、引用、常量引用

Can you explain to me the difference between returning value, reference to value, and const reference to value?

价值:

Vector2D operator += (const Vector2D& vector)
{
    this->x += vector.x;
    this->y += vector.y;
    return *this;
}

非常量引用:

Vector2D& operator += (const Vector2D& vector)
{
    this->x += vector.x;
    this->y += vector.y;
    return *this;
}

常量参考:

const Vector2D& operator += (const Vector2D& vector)
{
    this->x += vector.x;
    this->y += vector.y;
    return *this;
}

这样做有什么好处?我理解传递给函数的 const 引用背后的意义，因为您想确保不要修改引用指向函数内部的这个值.但是我对返回 const 引用的含义感到困惑.为什么返回引用比返回值好，为什么返回const引用比返回not-const引用好?

(v1 += v2) = v3;

v1 = (v2 += v3);