使用访问器公开类属性

发布于 2024-09-12 05:47:30 字数 460 浏览 7 评论 0原文

我不知道描述我的问题的正确技术术语，所以我举一个例子：

    private Point _PrivateVect = new Point();
    public Point Publicvect
    {
        get
        {
            return _PrivateVect;
        }
        set
        {
            _PrivateVect = value;
        }
    }

问题是，如果我想访问 Publicvect.X 我收到错误 Cannot edit 'Publicvect' 的返回值，因为它不是变量。有办法解决这个问题吗？或者我只需要永远执行 Publicvect = new Point(NewX, Publicvect.Y); ？

原文

I don't know the correct technical terms to describe my question, so I'll give an example:

    private Point _PrivateVect = new Point();
    public Point Publicvect
    {
        get
        {
            return _PrivateVect;
        }
        set
        {
            _PrivateVect = value;
        }
    }

The problem is that if I wanted to access Publicvect.X I get the error Cannot modify the return value of 'Publicvect' because it is not a variable. Is there a way around this? Or do I just need to do Publicvect = new Point(NewX, Publicvect.Y); forever?

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伪心 2024-09-19 05:47:30

可变结构是邪恶的另一个原因。一种解决方法是将维度公开为方便起见的访问器：

public Point PublicX {
    get {return _PrivateVect.X;}
    set {_PrivateVect.X = value;}
}
public Point PublicY {
    get {return _PrivateVect.Y;}
    set {_PrivateVect.Y = value;}
}

但除此之外；是的，您每次都需要执行new Point(x,y)，因为Point是一个结构体。当您通过属性访问它时，您将获得它的一个副本，因此，如果您改变该副本，然后丢弃该副本，您就会丢失更改。

Yet another reason that mutable structs are evil. One workaround is to expose the dimensions as accessors for convenience:

public Point PublicX {
    get {return _PrivateVect.X;}
    set {_PrivateVect.X = value;}
}
public Point PublicY {
    get {return _PrivateVect.Y;}
    set {_PrivateVect.Y = value;}
}

But other that this; yes you would need to do the new Point(x,y) each time, since Point is a struct. When you access it via a property you get a copy of it, so if you mutate the copy and then discard the copy, you simply lose the change.

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