类中的属性/方法是否占用内存中的空间？ [英] Do properties/methods in classes take space in memory?

问题描述

如果我们有这样的代码：

  public class Enemy 
 {
 public int hp; 
} 
  

然后一个敌人对象在32位机器中需要4个字节，在64位中需要8个字节（如果我错了，请纠正我）。

如果我们将其更改为如下所示：

  public class敌人
 {
 public int hp; 
 public void Attack（）{} 
} 
  

c> Enemy 对象仍然会采用与之前相同的内存量，对吗？

同样的：

  public class Enemy 
 {
 private int hp; 
 public int Hp {get {return hp; } set {hp = value; }} 
 
 $ / code> 

据我所知，属性是函数，但被当作变量，对吧？

所以，如果我们这样做了：

  public class Enemy 
 {
 public int Hp {set;得到; } 
} 
  

这是否意味着一个 Enemy 对象现在完全不占用内存空间？这是没有任何意义的。

甚至是这个，对于这个问题：

  public class Enemy 
 {
 public void DoSomething（）{} 
} 
  

有人可以解释吗？

解决方案

int 在C＃中始终是 System.Int32 ，无论32位还是64位应用程序，它总是占用4个字节的空间。

然而，对象中还有额外的开销。 Jon Skeet有一个博客文章，详细介绍了这里的一些内容，内存和字符串。

正如您所看到的，即使没有字段，以32位运行时，对象的基本大小也是12个字节。

然而，你是对的，因为具有代码的属性不一定会增加对象的大小。

然而，如果你使它成为一个自动属性，像这样：

  public int Hp {get;组;然后编译器会自动为你创建一个后台字段，以保存该属性的值，将再次占用空间。
 
If we had code like this:
public class Enemy
{
   public int hp;
}
Then an Enemy object would take 4 bytes in 32-bit machines, and 8 bytes in 64-bit (correct me if I'm wrong). 

If we change it to something like this:
public class Enemy
{
   public int hp;
   public void Attack() {}
}
An Enemy object would still take the same amount of memory as it did before, right?

The same for this:
public class Enemy
{
   private int hp;
   public int Hp { get { return hp; } set { hp = value; } }
}
From what I know, a property is function, but treated as a variable, right?

So if we did this:
public class Enemy
{
   public int Hp { set; get; }
}
Does that mean, an Enemy object now takes no memory space at all? That doesn't make any sense.

Or even this, for that matter:
public class Enemy
{
   public void DoSomething() { }
}
Can somebody explain?
解决方案
int in C# is always going to be System.Int32 which will always take up 4 bytes of space, regardless of 32-bit or 64-bit application.

However, there's additional overhead in an object. Jon Skeet has a blogpost that details some of this here, Of memory and strings.

As you can see, the base size of an object is 12 bytes when running as 32-bit, even if you have no fields.

You're right, however, in that a property that has code does not necessarily increase the size of the object.

However, if you make it an auto-property, like this:
public int Hp { get; set; }
Then the compiler will automagically create a backing field for you to hold the value of that property, which will again take up space.

                        
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