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

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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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