NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c++-part1/8386

Text File | 1996-08-05 | 2.2 KB | 74 lines

Path: piglet.cc.utexas.edu!not-for-mail
From: [e-mail]
Newsgroups: comp.lang.c++
Subject: Re: Memory allocation.
Date: 17 Feb 1996 10:56:00 -0600
Organization: The University of Texas at Austin
Message-ID: <4g51b0$[e-mail]>
References: <[e-mail]>
NNTP-Posting-Host: piglet.cc.utexas.edu

In article <[e-mail]>,
Kajdan Dimitry  <cerlpvk> wrote:
>I'm a starter so the question may seem stupid.

Not at all. I'm still a beginner to c++, so this was a good exercise
for me too.
8-)

>
>Would someone explain why this program works?
>
>int* func()
>{
>
>
> int b[10];
>   for(int i=0;i<9;i++)
>     b[i]=i;
>  return b;
>}

The fact that this particular function works is fortuitous. The problem is
that it returns a pointer to an array that is declared local to the
function. When the function returns, b[10] goes out of scope, and the
memory that was allocated for it is freed. So, the pointer returned by
func() doesn't point to a valid address anymore.
(I did not figure this out on my own - I gleaned this from another reply
to your post, and from a warning given me by my compiler when I compiled
this. Wish I could claim that i had, though 8-)).

>
>int main(){
>
> int* x= func();

Here, x points to a memory location that no longer belongs to b[10] in
func().

>
>The point is that b is allocated on the stack i.e without "new".

Well, when you declare an array as int b[10]; then memory for the array
is automagically allocated for it. you could also declare a pointer to
an array of 10 ints like this:

int *b = new int[10];

The scoping issues involved here are different, since new lets you allocate
memory on one function and then use or free it in another. memory allocated
with the new operator isn't tied to the lifetime of a function like
automatic variables such as "int b[10];" are.

So, return b will return a valid pointer if you declare b this way.

Other ways to fix func() are to declare the array globally, or you can
declare it static. In both cases, the array will 'persist' even after it goes
out of scope.

Of course, if I'm wrong about any of this, please correct me.
Thanks!

Lucien S.