-=> Quoting Neil Heller to Darin Mcbride
 NH> char aStr[10] = "abcdefgh";
 NH> char * aPtr = &aStr[5];
 
 NH> cout << aPtr << endl;
 
 DM> You want:
 
 DM> cout << (void*)aPtr << endl;
 NH> Just an honest question:  why, after aPtr was declared to be a pointer
I'd have a hard time believing this was a dishonest question.  ;-)
 NH> to char, would you need to cast it in order to print the address
 NH> (contents of the pointer).
You'd cast it to (void*) because, well, there is an operator for inserting
pointers that takes a void* as well as one for char*:
class ostream {
 //...
 ostream& operator<<(void*);
 ostream& operator<<(char*);
 //...
};
The first one prints out the address (since if it is a void*, you don't know
how to represent what it's pointing at).  The second, knowing what the
pointer is pointing at, prints out the array (C string) that it points at.
 NH> Instead of a C-style cast, would it be preferred in that situation to
 NH> do:
 NH> cout aPtr << end;
Not quite the same result.  ostream::operator<<(void*) usually prints out in
hex.  On 16-bit systems in large or huge memory model it will usually print
out the segment, too, in standard style (i.e. SSSS:OOOO).  int will usually
be in decimal (unless, of course, the ostream was previously informed
otherwise).  So you'd want the C++ style of:
cout (aPtr) << endl;
(Notice the ()'s - from what I can tell, they're required.)
Hope this helps,
... No. You cannot call 911. I am downloading my mail!
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