当它没有被定义时，它如何返回一个值？ [英] How is this returning a value when it has not been defined?

问题描述

我想了解重载操作符，我一直盯着这个时间比我想承认。我相信我理解类中的一切EXCEPT的运算符+成员。我试图教给自己与大量的信息，但我找不到任何信息，向我解释我在这里看到 - 我是一个坚定的信徒，如果我明白什么工作，然后我可以使用它更好。

I'm trying to understand overloading operators, and I've been staring at this for longer than I want to admit. I believe that I understand everything in the class EXCEPT the operator+ member. I'm trying to teach myself with the plethora of information available, but I can't find any information that explains to me what I'm seeing here--and I'm a firm believer if I understand HOW something works then I can use it better.

因此，大多数情况下，我的困惑在于编译器知道选择哪个变量temp。 （temp.x或temp.y）我意识到main（）是要求c.x和c.y，但是operator +似乎是返回未定义的东西。没有三元运算符或任何可以选择返回哪个运算符。

So, mostly, my confusion lies with how the compiler knows which variable of temp to choose. (temp.x or temp.y) I realize main() is asking for c.x and c.y, but operator+ seems to be returning something that wasn't defined yet. There is no ternary operator or anything that would make it pick which one to return.

#include <iostream>
using namespace std;

class CVector {
public:
    int x,y;
    CVector () {};
    CVector (int,int);
    CVector operator + (CVector);
};

CVector::CVector (int a, int b) {
    x = a;
    y = b;
}

CVector CVector::operator+ (CVector param) {
    CVector temp;
    temp.x = x + param.x;
    temp.y = y + param.y;
    return (temp);
}

int main () {
    CVector a (3,1);
    CVector b (1,2);
    CVector c;
    c = a + b;
    cout << c.x << "," << c.y;
    return 0;
}

推荐答案

因此，大多数情况下，我的困惑在于编译器如何知道要选择哪个
变量。

So, mostly, my confusion lies with how the compiler knows which variable of temp to choose.

不明白你的意思。编译器没有选择要返回的temp变量。 temp 是 CVector 类型的对象。它包含两个数据成员： x 和 y 。当使用此行创建时，这些成员作为 temp 的一部分存在：

I really don't understand what you mean by this. The compiler is not choosing a variable of temp to return. temp is an object of type CVector. It contains two data members, x and y. These members exist as part of temp when it is created with this line:

CVector temp;

然后，当你这样做：

return temp;

编译器没有任何选择。它返回整个对象，其中包括一个复合对象中的 x 和 y 。

There is nothing for the compiler to choose. It is returning the whole object, which includes both the x and the y in one compound object.

在您的主函数中，这行：

In your main function, this line:

c = a + b;

在 operator + > a 和 b 。然后将返回值（ temp ）（ operator = ）赋给 c 。因为你没有定义一个自定义赋值操作符，默认的赋值运算符只需从 temp 到 c 。因此， temp.x 分配给 cx 和 temp.y 分配给 cy 。

Calls operator+ on a and b. Then the return value (temp) is assigned (operator=) to c. Since you haven't defined a custom assignment operator, the default one kicks in, which simply does a memberwise assignment from temp to c. So, temp.x is assigned to c.x, and temp.y is assigned to c.y.

对于您的类，默认赋值运算符看起来像相同的操作语义）如果它被写出：

For your class, the default assignment operator would look like (or have identical operational semantics to) this, if it were written out:

CVector & CVector::operator=(const CVector & rhs)
{
    this->x = rhs.x;
    this->y = rhs.y;
    return *this;
}

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