sizeof类与int，函数，虚函数在C ++？ [英] sizeof class with int , function, virtual function in C++?

问题描述

这是一个在线C ++测试问题，已经完成。

This is an online C++ test question, which has been done.

#include<iostream>
using namespace std; 
class A
{

};
class B
{
int i; 
}; 

class C
{
void foo();
};
class D
{
virtual void foo();
};

class E
{
int i ; 
    virtual void foo();
};
class F
{
int i; 
    void foo();
};
class G
{
    void foo();
    int i;
    void foo1();
};

class H
{
    int i ;
    virtual void foo();
    virtual void foo1();
};
int main()
{
cout <<"sizeof(class A) : " << sizeof(A) << endl ;
cout <<"sizeof(class B) adding the member int i : " << sizeof(B) << endl ;
cout <<"sizeof(class C) adding the member void foo() : " << sizeof(C) << endl ;
cout <<"sizeof(class D) after making foo virtual : " << sizeof(D) << endl ;
cout <<"sizeof(class E) after adding foo virtual , int : " << sizeof(E) << endl ;
cout <<"sizeof(class F) after adding foo  , int : " << sizeof(F) << endl ;
cout <<"sizeof(class G) after adding foo  , int : " << sizeof(G) << endl ;
G g;
cout <<"sizeof(class G) after adding foo  , int : " << sizeof(g) << endl ;
cout <<"sizeof(class H) after adding int 2 virtual " << sizeof(H) << endl ;
return 0; 
}

输出：

sizeof(class A) : 1
sizeof(class B) adding the member int i : 4
sizeof(class C) adding the member void foo() : 1
sizeof(class D) after making foo virtual : 8
sizeof(class E) after adding foo virtual , int : 16
sizeof(class F) after adding foo  , int : 4
sizeof(class G) after adding foo   , unsigned int : 4
sizeof(class g) after adding foo  , unsigned int : 4
sizeof(class H) after adding int 2 virtual 16

我的问题：

为什么 siszeof（A）是1， sizeof（C）也是1？

Why siszeof(A) is 1 and sizeof(C) is 1 too ?

为什么 siszeof（H）是16，但 sizeof（G）是4？

Why siszeof(H) is 16 but sizeof(G) is 4 ?

为什么 siszeof（E）是16而 sizeof（F）是4？

Why siszeof(E) is 16 but sizeof(F) is 4 ?

为什么 siszeof（D）是8，但 sizeof（E）是16？

Why siszeof(D) is 8 but sizeof(E) is 16 ?

我的猜测：

虚函数是一个8字节的指针。
但是，我不知道为什么 E size是16？
将函数添加到空类不会更改其大小？

A virtual function is a pointer with 8 bytes. But, I do not know why E size is 16 ? Adding a function to an empty class does not change its size ?

任何帮助。

感谢

推荐答案

首先，虚函数不是指针有8个字节。在C ++中只有 sizeof（char）被保证为任意数量的字节。

First off, a virtual function is not a pointer with 8 bytes. In C++ nothing but sizeof(char) is guaranteed to be any number of bytes.

类中的虚函数增加它的大小（编译器依赖，但大多数 - 如果不是所有 - 它是这样的）。所有后续方法不会。非虚函数不会影响类的大小。

Second, only the first virtual function in a class increases its size (compiler-dependent, but on most - if not all - it's like this). All subsequent methods do not. Non-virtual functions do not affect the class's size.

这是因为类实例不保存指向方法本身的指针，而是指向一个虚函数表，每个类别一个。

This happens because a class instance doesn't hold pointers to methods themselves, but to a virtual function table, which is one per class.

所以，如果你有：

class A
{
   virtual void foo();
}

和

class B
{
   virtual void goo();
   virtual void test();
   static void m();
   void x();
}

你会有 sizeof（A）== sizeof （B）。

现在：

为什么siszeof（A）是1和sizeof（C）也是1？

Why siszeof(A) is 1 and sizeof(C) is 1 too ?

A 和 C 有大小1，因为它不允许一个类的大小为0.函数与它无关。

A and C have size 1 just because it's not allowed for a class to be of size 0. The functions have nothing to do with it. It's just a dummy byte.

为什么siszeof（H）是16而sizeof（G）是4？

Why siszeof(H) is 16 but sizeof(G) is 4 ?

G 只有一个用于记忆的成员 - int 。在你的平台上， sizeof（int）== 4 。 H 除了 int ，还有一个指向 vftable （虚函数表，见上文）。

G has only one member that accounts for memory - the int. And on your platform, sizeof(int) == 4. H, besides the int, also has a pointer to the vftable (virtual function table, see above). The size of this, size of int and allignment are compiler specific.

为什么siszeof（E）是16，而sizeof（F）是4 ？

Why siszeof(E) is 16 but sizeof(F) is 4 ?

以上解释 - 非虚拟方法不占用类中的内存。

Explained above - non virtual methods don't take up memory in the class.

为什么siszeof（D）是8，但sizeof（E）是16？

Why siszeof(D) is 8 but sizeof(E) is 16 ?

D 只包含 vftable 指针，在您的平台上显然是8个字节。 E 也有一个int， vftable 与8个字节对齐。所以它是这样的：

D only contains the vftable pointer which is apparently 8 bytes on your platform. E also has an int, and the vftable is aligned to 8 bytes. So it's something like:

class E

4 bytes for int |  4 padding bytes  |  8 bytes for vftable pointer  | 
| x | x | x | x |    |    |    |    | v | v | v | v | v | v | v | v |

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