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From: stevev@tekchips.UUCP (Steve Vegdahl)
Newsgroups: net.puzzle
Subject: Re: x to the x to the x ... (ANSWER/SPOILER)
Message-ID: <504@tekchips.UUCP>
Date: Sun, 12-May-85 17:10:20 EDT
Article-I.D.: tekchips.504
Posted: Sun May 12 17:10:20 1985
Date-Received: Tue, 14-May-85 23:43:34 EDT
References: <7219@watdaisy.UUCP> <3681@alice.UUCP> <481@tekchips.UUCP> <379@gitpyr.UUCP>
Organization: Tektronix, Beaverton OR
Lines: 47

> >                                               While it may be true that
> > most programming languages that use '^' for exponentiation define it to be
> > right-associative, I do not find either interpretation to be unreasonable.
> 
> The reason that right associativity is preferred is that:
> 
> 	(x ^ y) ^ z  is equivalent to  x ^ (y * z)
> 
> and the multiplication is certainly more efficient.  Therefore the only
> sensible way to associate exponentiation is to the right.  If a language
> does it the other way, then it's less than useless.

There is a great deal to be said for minimizing the number of exceptions
in a language.  If all operators except '^' are left-associative, many
would consider it say good language design to make '^' left-associative
as well.  I wonder how many people have used the left- or right-associativity
of an exponention operator in a program (i.e., written something of the form
x^y^z).  Is it worth making an exception to the language for a construct
that occurs so rarely?  Maybe, maybe not.

Which strikes you as cleaner language design?
	"All operators in language X are left-associative"
    or
	"All operators in language X are left-associative, except '^'"
    or
	"All operators in language X are left-associative, unless the
	amount on line 12b of Schedule A is less than 10% of adjusted
	gross income" :-)

ALGOL-60 and some of its descendants (e.g., SAIL), use '^' as a
left-associative exponentiation operator.  I stand by my earlier
statement that either interpretation is reasonable.  Remember that
precedence and left-(right-)associativity are properties of the
OPERATOR, not of the FUNCTION.  It is possible to define a language
with two exponentiation operators (of different precedence), one of
which is associates left the and other right.

I find the efficiency argument give above to be irrelavent.
If (x ^ y) ^ z is equivalent to x ^ (y * z), then an optimizing
compiler can transform the former to the latter.  I'm not even
sure that they are equivalent, when round-off error is considered.
Floating-point addition is NOT associative, for example.

		Steve Vegdahl
		Computer Research Lab.
		Tektronix, Inc.
		Beaverton, Oregon