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From: dave@murphy.UUCP (Rael's brother John)
Newsgroups: comp.lang.c
Subject: Re: Not only that, but...
Message-ID: <18@houligan.UUCP>
Date: Wed, 26-Nov-86 16:48:42 EST
Article-I.D.: houligan.18
Posted: Wed Nov 26 16:48:42 1986
Date-Received: Fri, 28-Nov-86 23:36:32 EST
Organization: Gould Electronics, Ft. Lauderdale, Florida.
Lines: 62

In article <1638@jade.BERKELEY.EDU>, mwm@eris.BERKELEY.EDU (Mike (Don't have 
strength to leave) Meyer) writes:

[person quoted is me]
>>Not only that, but: don't even *USE* an assigned-to object anywhere else
>>in the expression!
>
>There are operators in C that are guaranteed to
>preserve order, so you can make an assignment then use the variable
>assigned to.
>
>To wit, '||' and '&&' WILL be evaluated left-to-right (and short
>circuit), so you can safely write:
>
>	while ((c = getchar()) != TERMINATOR && c != EOF)
>		;

You're right.  Just to prove that I'm not dogmatic about it, I will reveal
an idiom of my own the violates my own assertion (the side effect involved
here is a function call rather than an assignment, but the principle is
the same):

       while ( (fgets(buffer,sizeof(buffer),foo) , !feof(foo) )
       {
              <stuff>
       }

I've found this to be a convenient way of setting up a loop to read lines
of text from and process them.  The test is clearly visible in the "while"
statement, no test-and-break within the loop is needed, and no "pump-priming"
read before the start of the loop is necessary.  This works because the
statement that you made about the && and || operators is also true of the
comma operator.  I think that where the person who posted the original
j = (k=1,k) + (k=2,k) business went wrong was in assuming that the
characteristics of the comma operator would apply globally across the +
operator, which of course isn't true of *any* operator in C.

I can think of one other case that seems like it should work, although I can't
prove it:

	j = ( (k=func())>0 ? k : 0)

The intent is that func is called and assigned to k, and then the value
assigned to j is either the value of k if k>0, or 0 if k<=0.  This seems
that it should be OK, because K&R states that for the question-mark
operator, only one of the second and third operands is evaluated, and
it seems to me that the first operand would first have to be fully evaluated in
order to determine which of the other two to evaluate.  But I can't find
any place where it comes right out and says this.

P.S.: how come there's no ^^ operator?
---
"I used to be able to sing the blues, but now I have too much money."
-- Bruce Dickinson

Dave Cornutt, Gould Computer Systems, Ft. Lauderdale, FL
UUCP:  ...!{sun,pur-ee,brl-bmd,bcopen}!gould!dcornutt
 or ...!{ucf-cs,allegra,codas}!novavax!houligan!dcornutt
ARPA: dcornutt@gswd-vms.arpa (I'm not sure how well this works)

"The opinions expressed herein are not necessarily those of my employer,
not necessarily mine, and probably not necessary."