Path: utzoo!utgpu!watmath!clyde!att!osu-cis!killer!ames!xanth!nic.MR.NET!umn-d-ub!rutgers!ucsd!ucbvax!hplabs!hpfcdc!hpislx!hplvli!boyne
From: boyne@hplvli.HP.COM (Art Boyne)
Newsgroups: comp.sys.ibm.pc
Subject: Re: Re: correct code for pointer subtraction (LONG reply)
Message-ID: <360005@hplvli.HP.COM>
Date: 4 Jan 89 16:04:42 GMT
References: <2245@iscuva.ISCS.COM>
Organization: Loveland Inst. Div
Lines: 39

chasm@killer.DALLAS.TX.US (Charles Marslett) writes:
> In article <142@bms-at.UUCP>, stuart@bms-at.UUCP (Stuart Gathman) writes:
>  > In article <18123@santra.UUCP>, tarvaine@tukki.jyu.fi (Tapani Tarvainen) writes:
>  > 
>  > > The same error occurs in the following program 
>  > > (with Turbo C 2.0 as well as MSC 5.0):
>  > 
>  > > main()
>  > > {
>  > >         static int a[30000];
>  > >         printf("%d\n",&a[30000]-a);
>  > > }
>  > 
>  > > output:  -2768
>  > 
>  > This is entirely correct.  The difference of two pointers is an *int*.
> 
> And unless you have a 15-bit computer, 30000 is a very representable *INT*,
> so please pay attention to the discussion before asserting something.  The
> compiler is generating a VERY WRONG ANSWER.

Ok guys, just hold it a minute and think: it only hurts for a little while.

Let's look at what the compiler is trying to do:  &a[30000]-a where a is an
integer array.  First, int's are 16-bit, ie., 2 bytes/int.  So, a[30000] is
located 60000 bytes away from a[0].  Ideally, the calculation should look like

      take address of a[30000]
      subtract address of a[0]
      divide by 2 to convert offset to integer pointer difference

But &a[30000]-a = 60000 bytes, too big for a 16-bit signed int.  Unsigned 60000
looks like -5536 signed.  Divided by 2 is -2768.  Voila.

You can argue that the compiler should use unsigned arithmetic for the pointer
calculation, or do a 32-bit calculation, but given the assumptions the compiler
operates under, the answer is reasonable (but not useful).

Art Boyne, boyne@hplvdz.HP.COM