Path: utzoo!attcan!uunet!decwrl!wuarchive!psuvax1!psuvm!cunyvm!ndsuvm1!mtus5!eacons
From: EACONS@MTUS5.BITNET (Ernie Anderson)
Newsgroups: comp.lang.c++
Subject: Whats wrong with this picture (or am I slow?)
Message-ID: <90281.014850EACONS@MTUS5.BITNET>
Date: 8 Oct 90 06:48:50 GMT
Organization: Computing Technology Services, Michigan Technological Univ.
Lines: 30

Hello all...
     I'm trying to pound the fundamentals of C++ through my thick skull,
and I've run into a stumper.  In "Programming in C++" by Dewhurst and
Stark in chapter 3 where they introduce their brain-dead example string
class, I have a question.  The code in question is
#include <stdio.h>
#include <string.h>
class String {
   char *str;
public:
   String() {str = new char[1]; *str = 0; }
   String(char *);
   void print() {printf(" %s", str); }
   ...
   };

Then later they do something like:
String a;
...
a = String (buffer);

Now when a was declared, a single character of storage was allocated, right?
and when buffer was converted to a String and assigned to a, the value of
str in that temporary instance for buffer was copied to a's str, right?  So
isn't the byte of storage allocated by declaring a lost?

Please either confirm this, or tell me where I'm going wrong...

Ernest J. Anderson
eacons@mtus5.cts.mtu.edu, eacons@mtus5.bitnet, ejanders@symmetry.cs.mtu.edu