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Text File | 1996-08-05 | 3.6 KB | 82 lines

Path: newshost.lanl.gov!tanmoy
From: [e-mail] (Tanmoy Bhattacharya)
Newsgroups: comp.lang.c
Subject: Re: Turbo C 3.0 - arithmetic in defines
Date: 13 Feb 1996 20:13:32 GMT
Organization: Los Alamos National Laboratory
Message-ID: <[e-mail]>
References: <4fqgkf$[e-mail]>
NNTP-Posting-Host: qcd.lanl.gov
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Content-Type: text
In-reply-to: [e-mail]'s message of 13 Feb 1996 17:09:35 GMT

In article <4fqgkf$[e-mail]> [e-mail] (Toby
K Hay) writes:
<snip>
   As a newcomer to C programming I'm having difficulty using arithmetic
   and #defined values to dimension arrays at compile time.  The following
   very short program reproduces my error:

Your error is very common misunderstanding amongst newcomers. I will
try to explain what I think is your problem.

   #define var1 16
   #define var2 485
   #define var3 ((var2-1)/var1)+1
   int main(void)
       {
       printf("\n%i    %i    %i    %i",var1,var2,var3,var1*var3);
       return 0;
       }

   Output from this is '16 485 31 481', not '16 485 31 496' as I should have
   expected.  That is, var3 is calculated correctly in its #define, but the

This is precisely your problem. A #define _never_ calculates
anythings: it simply defines a way of dumping in a sequence of tokens
when you need it. What the first define says is that when you later
use the token var1 in your code, and it is time for expanding the
token, it should be replaced by the token 16 instead. Thus, if you
wrote x+var1, the compiler would see token x, token +, token var1, and
then say "Aha! var1 is a `macro'!", and so replace var1 with 16. Then
it will say "Good! There is nothing more relevant to expand, so I have
found the three tokens x, + and 16, which means that the programmer
wants me to generate code for adding 16 to whatever x is." var2 is
similar.

Now see what happens when it sees var3? It says I have the sequence of
11 tokens on replacement: viz. ( ( var2 - 1 ) / ( var1 ) + 1, where I
have used spaces to show the separation between the tokens. Now, it
goes back and looks at the sequence and recognizes var2 and var1 as
macros, and replaces them further: ending up with ( ( 485 - 1 ) / ( 16
) + 1. It is only at this stage does it start interpreting this
expression: and either calculates it or compiles instruction for
calculating it. The result is of course 31.

So what happens when you say var1*var3? Well, if you follow through
the same argument, you find that you end up with 16*((485-1)/16)+1
(this time I amassuming you know what the tokens are, so I am not
showing the extraneous spaces that I was showing earlier) which it
then interprets, getting a 481, naturally.

   var1*var3 in the printf() statement is not calculated correctly.  Where
   am I going wrong?

Is it clear now? Remember to always define macros with an extra set of
parentheses whenever possible, unless they are one token long. Thus
#define var1 16 is fine (one token long),
#define complicated do { x:= 1; } while(0) is also fine (parentheses
not allowed), but #define var3 (((var2-1)/var1)+1) is what you ought
to do.

Cheers
Tanmoy
--
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