抽象类的大小

发布于 2024-08-28 10:18:49 字数 160 浏览 8 评论 0原文

如何找到抽象类的大小？

class A
{
    virtual void PureVirtualFunction() = 0;
};

由于这是一个抽象类，因此我无法创建此类的对象。如何使用“sizeof”运算符找到抽象类 A 的大小？

原文

How can I find the size of an abstract class?

class A
{
    virtual void PureVirtualFunction() = 0;
};

Since this is an abstract class, I can't create objects of this class. How will I be able to find the size of the abstract class A using the 'sizeof' operator?

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眼眸里的快感 2024-09-04 10:18:49

您可以使用sizeof运算符：

int a_size = sizeof(A);

You can use the sizeof operator:

int a_size = sizeof(A);

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你是年少的欢喜 2024-09-04 10:18:49

直接的答案是 sizeof 运算符可以采用对象或类型：

sizeof (A)

但实际上，为什么要这样做呢？你问这个问题的事实似乎令人担忧。

The direct answer is that the sizeof operator can take either an object or a type:

sizeof (A)

But really, why would you want to do this? The fact that you're asking this seems like cause for concern.

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一笔一画续写前缘 2024-09-04 10:18:49

如前所述，这个问题实际上没有意义——实例有大小，但类确实没有。由于无法创建抽象类的实例，因此它的大小基本上是一个毫无意义的概念。您可以使用 sizeof(A)，但结果并没有多大意义 - 由于空基类优化（举一个明显的例子），sizeof(A) 不一定告诉您使用A作为基类会对派生类的大小产生多大影响。

例如：

#include <iostream>

class A {};
class B {};
class C {};
class D {};

class Z : public A, public B, public C, public D {};

int main() { 
     std::cout << "sizeof(A) = " << sizeof(A);
     std::cout << "sizeof(Z) = " << sizeof(Z);
     return 0;
}

如果我在计算机上运行此命令，sizeof(A) 显示为 1。显而易见的结论是 sizeof(Z) 必须至少为 4 - - 但实际上，使用我手头的编译器（VC++、g++），它分别显示为 3 和 1。

As stated, the question doesn't really make sense -- an instance has a size, but a class really doesn't. Since you can't create an instance of an abstract class, its size is mostly a meaningless concept. You can use sizeof(A), but the result doesn't mean much -- thanks to the empty base class optimization (for one obvious example), sizeof(A) does not necessarily tell you how much using A as a base class will contribute to the size of a derived class.

For example:

#include <iostream>

class A {};
class B {};
class C {};
class D {};

class Z : public A, public B, public C, public D {};

int main() { 
     std::cout << "sizeof(A) = " << sizeof(A);
     std::cout << "sizeof(Z) = " << sizeof(Z);
     return 0;
}

If I run this on my computer, sizeof(A) shows up as 1. The obvious conclusion would be that sizeof(Z) must be at least four -- but in reality with the compilers I have handy (VC++, g++), it shows up as three and one respectively.

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