为什么类的大小只取决于数据成员而不是成员函数？ [英] Why class size depend only on data members and not on member functions?

问题描述

我想知道关于类大小的详细描述。
我想知道是否只有数据成员&成员函数没有任何虚拟关键字，那么为什么类大小仅取决于数据成员。
例如：

I want to know the detail description on the size of a class. I want to know if there is only data members & member function without any virtual keyword then why the class size depends only on data members. For an eg:

class A {
    int a;
public:
    int display() { 
    cout << "A=" << a << endl;
   }
};

当我检查 sizeof（A）我发现它是4字节。为什么会这样？为什么成员函数对类A的大小没有影响？

When I check the sizeof(A) i found that it is 4 byte. Why it is so? Why member function has no effect on the size of class A?

感谢

推荐答案

因为类的函数不保存在对象本身内。考虑到它在C编程方面，类 A 的每个函数都需要一个秘密参数， this 指针，因此实际上它们只是带有一个额外参数的函数。

Because the class's functions are not saved inside the object itself. Think of it in terms of C programming, every function of class A takes one secret parameter, the this pointer, so in actuality they are just functions with one extra parameter.

例如，假设它像这样：

int display(A* thisptr)
{
   //do something
   printf("%d",thisptr->a); 
   return;
}

因此，显示函数只保存一个简单的函数，

So the display function is saved as just a simple function with one extra parameter. The name though is mangled depending on the compiler.

我认为不同的规则适用于涉及函数指针的虚函数，但是我不确定也许别人可以启发我们就此事。

I believe that different rules apply for virtual functions which would involve function pointers but since I am not sure maybe someone else can enlighten us on this matter.

这篇关于为什么类的大小只取决于数据成员而不是成员函数？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！