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
 NH> to char, would you need to cast it in order to print the address
 NH> (contents of the pointer).
 There are many overloads for operator <<.  There is one for int's,
 one for char's, one for char*'s, etc.
 The overload for the char* is designed to handle zero terminated
 character arrays.  But there is also an overload defined to handle
 a void*.  That version will print the value of the pointer.
 NH> Instead of a C-style cast, would it be preferred in that situation to
 NH> do:
 NH> cout aPtr << end;
 I don't believe that will compile.  reinterpret_cast would compile.
 But you may well get a different output (thought the same value.)  On  
 my machine, the operator<<(void*) displays the answer in "pointer
 format" while operator<<(int) displays "plain old numbers".
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 þ Blue Wave/QWK v2.12 þ
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