c++ - union 成员(member)地址需要说明

Question

我写了下面的代码:

#include <iostream>
using namespace std;

union Packed
{
   char     i;
   short    j;
   int      k;
   long     l;
   float    f;
   double   d;
};

int main()
{
  cout<<"sizeof(Packed) = "<< sizeof(Packed)<<endl;
  Packed x;
  cout<<"Address j ="<<&(x.j)<<endl;
  cout<<"Address i ="<<&(x.i)<<endl;
  cout<<"Address k ="<<&(x.k)<<endl;
  cout<<"Address l ="<<&(x.l)<<endl;
  cout<<"Address f ="<<&(x.f)<<endl;
  cout<<"Address d ="<<&(x.d)<<endl;
  return 0;
}

输出:

sizeof(Packed) = 8

Address j =0x7fff587a71a0
Address i =

Address k =0x7fff587a71a0
Address l =0x7fff587a71a0
Address f =0x7fff587a71a0
Address d =0x7fff587a71a0

当我修改代码如下:

  cout<<"Address j ="<<(long)&(x.j)<<endl;
  cout<<"Address i ="<<(long)&(x.i)<<endl;
  cout<<"Address k ="<<(long)&(x.k)<<endl;
  cout<<"Address l ="<<(long)&(x.l)<<endl;
  cout<<"Address f ="<<(long)&(x.f)<<endl;
  cout<<"Address d ="<<(long)&(x.d)<<endl;

我得到输出:

Address j =140736718889408
Address i =140736718889408
Address k =140736718889408
Address l =140736718889408
Address f =140736718889408
Address d =140736718889408

我有点困惑，为什么当我只是打印 union 中所有变量的基地址时，我没有得到 union 中字符变量的正确地址，这应该与“长”时确认的相同在代码的第二部分进行类型转换。

请为混淆提供解释。输出列为在 Fedora FC17 Linux 机器上获得。

Answer 1

在 ostream 中有一个 char const* 变量的重载。将所有指针转换为 void const* 以获得正确的结果。

哦，在我忘记之前:使用 static_cast() 并且永远不要使用 C 风格的转换，除非你真的知道你必须这样做!