SIMPLE malloc & pointer question

[Richard Tobin]

In article <7206 at helios.TAMU.EDU> jdm5548 at 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 at uk.ac.ed             
AI Applications Institute,           ARPA:  R.Tobin%uk.ac.ed at nsfnet-relay.ac.uk
Edinburgh University.                UUCP:  ...!ukc!ed.ac.uk!R.Tobin