在Visual Studio中C ++指向类数据成员的指针的地址

本文介绍了在Visual Studio中C ++指向类数据成员的指针的地址的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我在阅读这本书 C ++对象模型。在书中有一个例子：

I'm reading the book Inside the C++ Object Model. In the book there's an example like:

struct Base1 { int v1; }; struct Base2 { int v2; }; class Derived : public Base1, public Base2 {}; printf("&Derived::v1 = %p\n", &Derived::v1); // Print 0 in VS2008/VS2012 printf("&Derived::v2 = %p\n", &Derived::v2); // Print 0 in VS2008/VS2012

在上面的代码中，打印地址Derived :: v1 & Derived :: v2将都是 0 。但是，如果通过变量打印相同的地址：

In the previous code, the print of address Derived::v1 & Derived::v2 will both be 0. However, if print the same address via a variable:

int Derived::*p; p = &Derived::v1; printf("p = %p (&Derived::v1)\n", p); // Print 0 in VS2008/VS2012 as before p = &Derived::v2; printf("p = %p (&Derived::v2)\n", p); // Print 4 in VS2008/VS2012

通过检查& Derived :: v1和p

By examining the size of &Derived::v1 and p, I get 4 in both.

// Both are 4 printf("Size of (&Derived::v1) is %d\n", sizeof(&Derived::v1)); printf("Size of p is %d\n", sizeof(p));

Derived :: v1的地址为 0 ，但地址of Derived :: v2将为 4 。我不明白为什么& Derived :: v2在分配给变量时变为 4 。

The address of Derived::v1 will be 0, but the address of Derived::v2 will be 4. I don't understand why &Derived::v2 became 4 when assign it to a variable.

检查汇编代码，查询Derived :: v2的地址，它被翻译成 0 ;但是当它分配给一个变量时，它被转换为一个 4 。

Examine the assembly code, when directly query the address of Derived::v2, it is translated to a 0; but when assign it to a variable, it gets translated to a 4.

我在VS2008& VS2012，结果是一样的。所以我认为必须有一些理由让微软选择这样的设计。

I tested it on both VS2008 & VS2012, the result is the same. So I think there's must be some reason to make Microsoft choose such design.

如果你这样做：

d1.*(&Derived::v2) = 1;

显然& Derived :: v2 不是 0 。为什么编译器区分这两种情况？

Apparently &Derived::v2 is not 0. Why does the compiler distinguish this two cases?

任何人都可以告诉后面发生的事情吗？谢谢！

Can anyone please tell the thing happens behind? Thank you!

- 编辑 -

对于那些认为& Derived :: v1 doesn'得到有效的地址。

For those think the &Derived::v1 doesn't get a valid address. Haven't you ever did this?

Derived d1, d2; d1.*p = 1; d2.*p = 1;

推荐答案

海报问我这个，也怀疑类似的错误原因。这不是特定于VC ++。

The poster asked me about this, and at first I also suspected similar wrong causes. This is not specific to VC++.

事实证明发生的是，& Derived :: v2 不是 int Derived :: * ，但 int Base2 :: * 零，因为它是相对于Base2的偏移。当你显式地将它转换为 int Derived :: * 时，偏移量被更正。

It turns out that what's happening is that the type of &Derived::v2 is not int Derived::*, but int Base2::*, which naturally does have an offset of zero because it's the offset with respect to Base2. When you explicitly convert it to an int Derived::*, the offset is corrected.

VC ++或GCC或Clang ...我正在使用stdio / printf作为海报使用。

Try this code on VC++ or GCC or Clang... I'm sticking with stdio/printf as the poster was using.

struct Base1 { int a; }; struct Base2 { int b; }; struct Derived : Base1, Base2 { }; #include <cassert> #include <cstdio> #include <typeinfo> using namespace std; int main () { printf( "%s\n", typeid(&Derived::a).name() ); // mentions Base1 printf( "%s\n", typeid(&Derived::b).name() ); // mentions Base2 int Derived::* pdi = &Derived::b; // OK int Base2::* p2i = &Derived::b; // OK //int Base1::* p1i = &Derived::b; // ERROR assert( sizeof(int*) == sizeof(pdi) ); printf( "%p %p", p2i, pdi ); // prints "(nil) 0x4" using GCC 4.8 at liveworkspace }