Path: utzoo!attcan!uunet!munnari.oz.au!comp.vuw.ac.nz!cavebbs!frank
From: frank@cavebbs.gen.nz (Frank van der Hulst)
Newsgroups: comp.lang.c
Subject: Re: Why declare returned pointers static?
Message-ID: <1991Mar01.081914.13539@cavebbs.gen.nz>
Date: 1 Mar 91 08:19:14 GMT
References: <1991Feb25.074222.12846@msuinfo.cl.msu.edu> <484@bria>
Organization: The Cave MegaBBS, Public Access Usenet, Wellington, NZ
Lines: 38

In article <484@bria> uunet!bria!mike writes:
>In an article, draper@galaxy.cps.msu.edu (Patrick J Draper) writes:
>|DATA_STRUCTURE *foo ()
>|{
>| static DATA_STRUCTURE *bar;
>|
>|   bar = (DATA_STRUCTRE *) malloc (sizeof (DATA_STRUCTRE));
>|   return (bar);
>|}
>|
>|and called like this:
>|DATA_STRUCTRE *bar;
>|
>|   bar = foo ();
>|
>|My question is: Why does this work? Is the malloc'd space static and not
>|the pointer? From my tests with Turbo C++, with a static pointer, the
>|malloc'd space never gets overwritten accidentally. With a normal
>|pointer, the malloc'd space can be corrupted by things like another
>|malloc, etc.
>
>Memory that has been allocated ala malloc() is global within the context
>of the process allocating it.  The pointer is not.  Remember, when you
>declare a pointer, you *are* declaring storage (for the address).  If the
>pointer is an automatic variable, the memory that contains the address
>is released unless it is declared static.  Your chunk of memory that you
>grabbed with malloc() is still there; the pointer to it is not.
>
But the contents of the pointer (i.e. the address of the malloc'ed memory)
is being returned (presumably via CPU registers) to be stored in ably)
yet another DATA_STRUCTURE * variable. So it doesn't matter if the bar
variable inside foo() is lost. Unless TC++ is attempting some garbage
collection at the end of foo(), and doesn't realise that the memory is
still being used, and pointed to in the CPU registers.
 
-- 

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