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From: levy@ttrdc.UUCP (Daniel R. Levy)
Newsgroups: comp.lang.c
Subject: Re: condition expressions
Message-ID: <1694@ttrdc.UUCP>
Date: Thu, 14-May-87 22:49:03 EDT
Article-I.D.: ttrdc.1694
Posted: Thu May 14 22:49:03 1987
Date-Received: Sun, 17-May-87 00:29:08 EDT
References: <1130@ius2.cs.cmu.edu> <148@nbisos.NBI.COM> <874@cc5.bbn.com.BBN.COM>
Organization: AT&T, Skokie, IL
Lines: 38
Summary: is this explanation REALLY clear?

In article <874@cc5.bbn.com.BBN.COM>, keesan@cc5.bbn.com.BBN.COM (Morris M. Keesan) writes:
< In article <801@rtech.UUCP> daveb@rtech.UUCP (Dave Brower) writes:
< <Which reminds me of a general problem with the conditional operator.
< <What is the type of the result of evaluating:
< <	(expr) ? (double) x : (char)y
< <I believe it is undefined.   One could adopt a number of possible
< <rules for interpretation:
< <           . . . [possible rules omitted for brevity]
< <I'm inclined to call it an error.  Thoughts, anyone?
<  The Draft Proposed American National Standard is even clearer:
< "If both . . . have arithmetic type, the usual arithmetic conversions are
< performed to bring them to the same type and the result has that type."
< 
< In the example given, the result of the expression is either (double)x or
< (double)(int)(char)y, following the order of conversions specified by K&R.
< (Semantics of conversions is not specified as precisely in the DPANS.)

Well, this quotation says that the "usual arithmetic conversions" are
performed to bring the expressions to the same type, but it still doesn't
say WHAT type prevails if the two differ; "usual arithmetic conversions"
is begging the question!  (Otherwise, why not quote some other part of the
spec that says what "usual arithmetic conversions" are with respect to members
of tertiary expressions ... oh boy, here we recurse....?)  Why "double" in
the example given: is it supposed to be because double is the type of the
first subexpression, or is it supposed to be because double can accommodate
a char but not vice versa, or what?  This is a valid issue in the case where
there is no lvalue involved:

double x; char y; int i;
...
printf("%lf\n",
/* why wouldn't that be "%c\n", or even (i ? "%lf\n" : "%c\n") ? */
i ? x : y);
-- 
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