为什么不显示char数据的地址? [英] Why is address of char data not displayed?

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 << "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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