为什么将char指针流传输到cout不能打印地址？

Question

When I print a char pointer with printf() , it makes the decision with conversion specifier whether the address should be printed or the whole string according to %u or %s.

But when I want to do the same thing with cout , how will cout decide what should be printed among address and whole string? Here is an example source:

int main()
{
  char ch='a';
  char *cptr=&ch;
  cout<<cptr<<endl;
  return 0;
}

Here, in my GNU compiler, cout is trying to output ch as a string.

How I can get address of ch via cptr using cout ?

Answer 1

Overload resolution selects the ostream& operator<<(ostream& o, const char *c); which is used for printing C-style strings. You want the other ostream& operator<<(ostream& o, const void *p); to be selected. You are probably best off with a cast here:

 cout << static_cast<void *>(cptr) << endl;

Answer 2

cout prints a string if it receives a char * , simple as that.

Here are the overloads for operator << for ostream :

ostream& operator<< (bool val);
ostream& operator<< (short val);
ostream& operator<< (unsigned short val);
ostream& operator<< (int val);
ostream& operator<< (unsigned int val);
ostream& operator<< (long val);
ostream& operator<< (unsigned long val);
ostream& operator<< (float val);
ostream& operator<< (double val);
ostream& operator<< (long double val);
ostream& operator<< (const void* val);

ostream& operator<< (streambuf* sb);

ostream& operator<< (ostream& ( *pf )(ostream&));
ostream& operator<< (ios& ( *pf )(ios&));
ostream& operator<< (ios_base& ( *pf )(ios_base&));

ostream& operator<< (ostream& out, char c );
ostream& operator<< (ostream& out, signed char c );
ostream& operator<< (ostream& out, unsigned char c );


//this is called
ostream& operator<< (ostream& out, const char* s );
ostream& operator<< (ostream& out, const signed char* s );
ostream& operator<< (ostream& out, const unsigned char* s );

If you want the address, you want:

ostream& operator<< (const void* val);

so you need to cast to const void* .

Answer 3

I would just cast it to a void* so it doesn't try to interpret it as a C-string:

cout << (void*) cptr << endl;

However, a safer option would be to use static_cast as in dirkgently's answer (that way the cast is at least checked at compile time).

Answer 4

As Luchian said, cout knows what to print based on the type. If you want to print the pointer value, you should cast the pointer to void* which will be interpated as a pointer.