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From: sjc@key.COM (Steve Correll)
Newsgroups: comp.lang.fortran
Subject: Re: Side effects in boolean expressions
Message-ID: <2041@key.COM>
Date: 14 Aug 90 23:32:58 GMT
References: <2022@key.COM> <59110@lanl.gov>
Organization: Key Computer Labs, Fremont, CA
Lines: 40

In article <59110@lanl.gov>, jlg@lanl.gov (Jim Giles) writes:
> From article <2022@key.COM>, by sjc@key.COM (Steve Correll):
> > In article <58237@lanl.gov>, jlg@lanl.gov (Jim Giles) writes:
> >> > 	X .GT. Y .OR. L(Z)
> > [...]
> > C is predictable. The Fortran standard says that the processor _need_ _not_
> > execute the function L if X exceeds Y, and that therefore the program must
> > assume Z is undefined. The C standard for "(x > y) || l(&z)" says that the
> > processor _must_ _not_ execute l if x exceeds y...
> 
> C is predictable only if you know in advance whether X was greater than
> Y.  If I knew that, I wouldn't have tested it.  This is what I meant
> when I said that neither language is defined in such a way as to be
> able to tell a priori (from the first) whether the function would be
> evaluated or not...
> It also means that I can't
> apply the mathematical properties of the 'or' operator to correctness
> proofs of the program.

Perhaps we simply define "predictable" and "a priori" differently, but
consider the following. Even though you don't know the value of x in advance,
the C code is legal and its behavior predictable; the Fortran is neither:

	if ((x < 0.0) || (sqrt(x) < 10.0)) ...
	IF ((X .LT. 0.) .OR. (SQRT(X) .LT. 10.0)) ...

If x is negative, a legal C processor will not call sqrt, but a legal Fortran
processor may or may not--in fact, it could vary with the phase of the moon!
By refusing to define the order of evaluation, Fortran gives the processor
more freedom (e.g. to optimize).

When proving correctness of a C program, notice that:

	if (p0 || p1) s0;

is equivalent to:

	if (p0) s0; else if (p1) s0;
-- 
...{sun,pyramid}!pacbell!key!sjc 				Steve Correll