本文介绍了为什么不显示char数据的地址?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
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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