passing variable numbers of arguments

[David MacKenzie]

The following program produces unexpected results:

#include <varargs.h>

main ()
{
  printf ("foo:\n");
  foo (1, 2, 3, 0);
  printf ("bar:\n");
  bar (4, 5, 6, 0);
}

foo (va_alist)
  va_dcl
{
  bar (va_alist);
}

bar (va_alist)
  va_dcl
{
  va_list list;
  int i;

  va_start (list);
  while (i = va_arg (list, int))
    printf ("%d\n", i);
  va_end (list);
}

On a 68000 machine, it produces:

foo:
1
14679744
160
1
2
3
bar:
4
5
6

and on a VAX it produces:
foo:
1
bar:
4
5
6

whereas the output I wanted is:

foo:
1
2
3
bar:
4
5
6

In other words, if I call bar () directly, it works, but if I call it via
foo (), it breaks, in an implementation-dependant way, no less.  Why?

Is there anyway to get this to work?  The problem has arisen as I
am trying to port Allan Holub's (DDJ, April 1988) integer-only
printf from <stdarg.h> to <varargs.h>.  printf, sprintf, and fprintf
are small little functions that call a function called idoprnt and both the
small front-end functions and idoprnt seem to need to accept a variable
number of arguments.

David MacKenzie
edf at rocky2.rockefeller.edu
---
p.s. I post from this account because inews doesn't seem to be able to post
new articles on rocky2 -- very strange, since it can post followups just
fine.