Hi,

How do you pass an arbitrary array of pointers to a function such that
they can each be assigned to something so that the caller can observe
these assignments after the call? My code below is one of many attempts
to get a handle on how this is done however in this particular version I
get this error from gdb:

Program received signal SIGSEGV, Segmentation fault.
0x804841c in getlist (list=0xbffff8c0) at t10.c:13
13              *list[1] = n2;
(gdb)

Thanks as usual,

Michael B. Allen

/* t10.c */

#include <stdlib.h>
#include <stdio.h>

char *n1 = "fred";
char *n2 = "barney";

int
getlist(void ***list)
{
    *list[0] = n1;
    *list[1] = n2;
    return 2;

int
main()
{
    int n = 2;
    char **names = malloc(sizeof *names * n);
    n = getlist((void ***)&names);
    printf("%s\n", names[0]);
    printf("%s\n", names[1]);
    return 0;

Sent via Deja.com http://www.*-*-*.com/
Before you buy.

`void' here is bogus.  You should use `char'.

This is equivalent to *(list[0]), which is equivalent to **(list
+ 0).  Okay, but abstractionally challenged.

This is equivalent to *(list[1]), which is equivalent to **(list
+ 1).  *BOOM*  

You mean (*list)[1], which is equivalent to *((*list) + 1).

Drop the cast.

By the way, there's no need for the triple pointer here.  You
could do it just as well, and with less confusion, with a double
pointer.  You only need the triple pointer if you're going to
include something like
        *names = malloc (sizeof *names * n);
within getlist() itself.
--
"You call this a *C* question? What the hell are you smoking?" --Kaz