为什么不显示字符数据的地址？

问题描述

  class Address {
 int i; 
 char b; 
 string c; 
 public：
 void showMap（void）; 
}; 
 
 void Address :: showMap（void）{
 cout< address of int：<< & i< endl; 
 cout<< 地址char：< & b< endl; 
 cout<< string of string：<< & c< endl; 
} 
  

输出为：

 地址int：something 
 char的地址：//没有，空白区域，没有显示
字符串地址：something 
  
 
 
 
 为什么？ 
 
 
 
事情：如果int，char，string是在public，那么输出是
  ... int：something 
 ... char：
 ... string：something_2 
  
  something_2  -  something 总是等于8. 为什么？ （不是9） 
解决方案
当您使用b的地址时， char * 。 运算符<< 将其解释为C字符串，并尝试打印字符序列而不是其地址。
 
 
 
 try  cout<< 地址char：< （void *）& b< 
 
 
 
 像Tomek所说，在这种情况下使用更合适的转换是 static_cast ，这是一个更安全的替代品。这是一个使用它而不是C风格的转换的版本：
  cout< 地址char：< static_cast< void *>（& b）<< endl; 
  
 
class Address {
      int i ;
      char b;
      string c;
      public:
           void showMap ( void ) ;
};

void Address :: showMap ( void ) {
            cout << "address of int    :" << &i << endl ;
            cout << "address of char   :" << &b << endl ;
            cout << "address of string :" << &c << endl ;
}
The output is:
         address of int    :  something
         address of char   :     // nothing, blank area, that is nothing displayed
         address of string :  something
Why?
Another interesting thing: if int, char, string is in public, then the output is
  ... int    :  something
  ... char   :
  ... string :  something_2
something_2 - something is always equal to 8. Why? (not 9)
 解决方案 
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char   :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char   :" << static_cast<void *>(&b) << endl;
                        
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