空访问器重要吗？关于值类型及其修改

发布于 2024-08-27 13:42:41 字数 560 浏览 10 评论 0原文

由于“a”是键入的值，我的以下代码不起作用。但我认为即使没有访问器它也不起作用，但它确实起作用：

class Program
    {
        a _a  //with accessors it WONT compile
        {
            get; 
            set;
        }
        static void Main(string[] args)
        {
            Program p = new Program();
            p._a.X = 5; //when both accessors are deleted, compiler does not
                        //complain about _a.X not being as variable
        }
    }
    struct a
    {
       public int X;
    }

它不起作用，因为“a”是结构。但是当我从“_a”实例中删除访问器时，它就起作用了。我不明白为什么。谢谢

原文

I have following code that does not work due to "a" being a value typed. But I thought it would not work even without accessors, but it did:

class Program
    {
        a _a  //with accessors it WONT compile
        {
            get; 
            set;
        }
        static void Main(string[] args)
        {
            Program p = new Program();
            p._a.X = 5; //when both accessors are deleted, compiler does not
                        //complain about _a.X not being as variable
        }
    }
    struct a
    {
       public int X;
    }

It does not work as "a" is struct. But when I delete accessors from "_a" instance, it works. I do not understand why.
Thanks

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悲欢浪云 2024-09-03 13:42:41

值类型的主要特征是它们是复制的而不是通过引用传递的。

当您有一个值类型和一个访问器时，您本质上有一个从方法返回的值类型，这会导致复制（以下两个示例是相同的）：

ValueType Property { get { return x; } } // Will make a copy of x
ValueType Method() { return x; }    // Will make a copy of x

如果您现在分配给返回的值，则您正在分配x 的副本。因此，对该属性返回的值所做的任何更改都将立即丢失。

当您删除 { get; 时} 访问器，你现在有了一个基本字段，例如：

int field;

ValueType field;

这意味着没有复制，这意味着当分配给该字段时，你不再分配给副本。

The main feature of value types is that they are copied rather than being passed by reference.

When you have a value type, and an accessor, you essentially have a value type being returned from a method, which causes a copy (the following two examples are the same):

ValueType Property { get { return x; } } // Will make a copy of x
ValueType Method() { return x; }    // Will make a copy of x

If you now assign to the returned value, you're assigning to a copy of x. So any changes made to the value returned from the property will be immediately lost.

When you remove the { get; } accessor, you've now got a basic field, e.g.:

int field;

ValueType field;

Which means no copy is made, which means when assigning to the field, you're no longer assigning to a copy.

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維他命╮ 2024-09-03 13:42:41

p._a.X = 5; 不起作用的原因是 p._a 返回 a 类型的值。价值观不能改变。但如果将值放入变量中，则可能会更改变量的值。

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彡翼 2024-09-03 13:42:41

您无法删除这两个访问器。

这样：

a _a;

它可以工作，但它不再是一个属性。

编辑：对于属性，从 p._a 获得的值是函数调用的结果。即使你修改了它，修改后的值也绝不会“写回”到“原始”_a。相反，您只需修改 getter 函数返回的临时值。

C# 可以允许这样做，但这会导致混乱，因为人们会期望在 p._a.X = 5; 之后int xx = p._a.X;xx 的值将为 5。但事实并非如此。因为 p_.a 确实不是一个变量:-)

不同的是，

a _a;

你的 _a 是一个字段；如果

a _a { get; set; }

_a 是一个属性。而且这种情况

a _a { }

是不允许的。

You can't delete both accessors.

This way:

a _a;

it works, but it's not a property any more.

Edit: With a property, the value you get from p._a is a result of a function call. If you even modify it, the modified value will by no means "written back" to the "original" _a. Instead, you will just modify a temporary, returned by the getter-function.

C# could allow this, but it would lead to confusion, since people would expect that after p._a.X = 5; int xx = p._a.X; the value of xx would be 5. But it won't be so. Because p_.a is indeed not a variable :-)

The difference is that with