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From: GHGAQZ4@cc1.kuleuven.ac.be
Newsgroups: comp.sys.amiga.programmer
Subject: Re: Compiler code (was a flame fest) (now a lesser flame fest)
Message-ID: <91113.092907GHGAQZ4@cc1.kuleuven.ac.be>
Date: 23 Apr 91 09:27:07 GMT
References: <1991Apr18.212939.3461@kessner.denver.co.us>
 <1991Apr19.032052.26387@engin.umich.edu>
 <91112.093750GHGAQZ4@cc1.kuleuven.ac.be>
 <1991Apr23.020319.7216@engin.umich.edu>
Organization: K.U.Leuven - Academic Computing Center
Lines: 46

>>
>>I think you are wrong. Look at the following results :
>>I have tried this with our mainframe compiler :
>>   int n=4;
>>
>>   (n--)*(n++)   ->   12
>>   (n++)*(n--)   ->   20
>>   (--n)*(++n)   ->   16
>>   (++n)*(--n)   ->   16
>>   (--n)*(n++)   ->   12
>>   (++n)*(n--)   ->   20
>>   (n--)*(++n)   ->   16
>>   (n++)*(--n)   ->   16
>>
>>          Jorrit Tyberghein
>
>  You missed my point.  I was saying that ++ and -- are unary operators which
>will be evaluated before the multiplication.  You've pointed out the argument
>this stems from, which is that there's no way to know which side of the mult-
>iplication is evaluated (You're compiler does the left first..).

There is something I don't understand about this. You say that the compiler
evaluates left first. But how do you interprete (--n)*(++n) ? If the compiler
would go from left to right he should first compute --n (or n=n-1). This is
the left side of the expression. And after that the compiler would evaluate
(++n) which is the right side of the expression. Obviously I'm thinking wrong
because this would give 12 instead of 16. What really happens here is that
   (--n)*(++n) gets translated to :
   --n
   ++n
   (n)*(n)   (or something equivalent)
I've looked at the compiled code and this is indeed what is evaluated.
But why does (n++)*(n--) not evaluate to the equivalent :
   (n)*(n)
   n++
   n--
I thought that ALL --x or ++x operators where to be evaluated BEFORE the
evaluation of the total expression happens (and this is presumably also
what happens) and that ALL x++ or x-- operators where to be evaluated
AFTER the evaluation of the total expression (and this is presumably
not true). This would seem more logical to me.

By the way. Lattice C generates exactly the same results.


                          Jorrit Tyberghein