Path: utzoo!utgpu!attcan!uunet!lll-winken!lll-tis!helios.ee.lbl.gov!pasteur!ames!hc!lanl!jlg
From: jlg@lanl.gov (Jim Giles)
Newsgroups: comp.lang.fortran
Subject: Re: Fortran versus C for numerical analysis
Message-ID: <3182@lanl.gov>
Date: 2 Sep 88 19:03:45 GMT
References: <3534@s.cc.purdue.edu>
Organization: Los Alamos National Laboratory
Lines: 38

From article <3534@s.cc.purdue.edu>, by ags@s.cc.purdue.edu (Dave Seaman):
> 
> On the contrary, Fortran's parentheses control only grouping and have
> nothing to do with evaluation order.  Consider the statement
> 
> 	Y = A(X) + (B(X) + C(X))
> 
> where A, B and C are external functions.  There are 3! = 6 possible orders
> of evaluation for the three functions.  All six of them are legal in
> Fortran.  For example, a perfectly legal code sequence would be the
> following:
> 
> 	1. Evaluate C(X) and place the result in a temporary, t1.
> 	2. Evaluate A(X) and place the result in a temporary, t2.
> 	3. Evaluate B(X) and place the result in a temporary, t3.
> 	4. Add t3 to t2 and place the result in t3.
> 	5. Add t3 to t1 and store the result in Y.

This is simply _not_ true.  Read the standard before claiming things
like this.  ANSI X3.9-1978 FORTRAN 77, page 6-18, line 30:
   
    In addition to the parenthesis required to establish the
    desired interpretation, parenthesis may be included to
    restrict the alternative form that may be used by the
    processor in the actual evaluation of the expression. This
    is useful for controlling the magnitude and accuracy of
    intermediate values developed during the evaluation of an
    expression.  For example, in the expression

              A+(B-C)

    the term (B-C) must be evaluated and then added to A. [...]

This seems pretty clear.  To be sure, the three function calls in your
example can be evaluated in any order.  But, the result of functions
B and C _must_ be added first, then added to the result of function A.

J. Giles
Los Alamos