Path: utzoo!utgpu!news-server.csri.toronto.edu!mailrus!uunet!mcsun!ukc!edcastle!aiai!richard
From: richard@aiai.ed.ac.uk (Richard Tobin)
Newsgroups: comp.lang.c
Subject: Re: SIMPLE malloc & pointer question
Message-ID: <3143@skye.ed.ac.uk>
Date: 7 Aug 90 14:22:03 GMT
References: <7206@helios.TAMU.EDU>
Reply-To: richard@aiai.UUCP (Richard Tobin)
Organization: AIAI, University of Edinburgh, Scotland
Lines: 35

In article <7206@helios.TAMU.EDU> jdm5548@diamond.tamu.edu (James Darrell McCauley) writes:

>  inita(a,b);
>  printf("main(): a[2]=%d\n",a[2]);

>inita (a,b)
>int a[],b[];
>{
>  a=(int *)malloc( (unsigned) 4*sizeof(int));
>  a[2]=3;

Arguments in C are passed by value, not reference.  inita() gets a copy
of main()'s variable "a", so when inita() returns the value in main()
is unchanged.

If you want to modify a variable in the parent procedure, you should
pass a pointer to it:

  inita(&a,b);

and declare the argument appropriately:

  int **a;			/* a is a pointer to a pointer to integers */

and assign to what the argument points to, not the argument itself

  (*a)=(int *)malloc( (unsigned) 4*sizeof(int));
  (*a)[2]=3;

-- Richard

-- 
Richard Tobin,                       JANET: R.Tobin@uk.ac.ed             
AI Applications Institute,           ARPA:  R.Tobin%uk.ac.ed@nsfnet-relay.ac.uk
Edinburgh University.                UUCP:  ...!ukc!ed.ac.uk!R.Tobin