Re: OASIS: Infix Operator "^"

[John Machin <sjmachin lexicon net>]

On 23/02/2007 6:09 AM, Leonard Mada wrote:
> Chapter 6.3.5 of the OASIS OpenFormula document (version 2007-02-14) 
> describes the Infix Operator (^).
>
> However, it contains a lot of mathematical ERRORS:
[snip]
> 2. The classic ERROR: -4^2
> Mathematically, this yields always -(4^2)=-16, NEVER +16!!!
> There is NO unary operator in mathematics that converts a number to its 
> negative, like described in the documents. I would suggest consulting a 
> dedicated mathematical program. As stated in issue 
> http://www.openoffice.org/issues/show_bug.cgi?id=66735 on the OOo website:
> *Mathematica* v5.(x) returns -16
> *R* v2.4.0 returns -16
> other software (like Scilab, maxima, octave, as pointed in that Calc 
> issue) report -16, too.
>
> gnumeric takes an interesting approach: it converts the first expression 
> to =(-4)^2, which indeed is then +16. So, this last is mathematically 
> correct and, at the same time, users who know a bit more about 
> mathematics, recognize that gnumeric did NOT calculated what they 
> expected. (it is transparent what has happened)
>
> Stating however, that -4^2 = +16 is plain wrong!!! And inventing various 
> mathematical rules that break the old ones is really NOT good for a 
> standard.
>
> I hope this gets corrected.

IMextremelyHO: Anyone who attempts to rely on knowing what the 
precedence of operators is in any language -- apart from () */ and +- -- 
instead of putting () in the expression is just plain crazy.

It is a well known fact that language designers make arbitrary decisions 
about operator precedence. One of the more ludicrous cases is in Pascal 
and its inheritors: the () in "if (a < b) and (c < d)" are *necessary* 
Why? Because without the (), that would be interpreted as "if a < (b and 
c) < d", which would be valid only if all 4 variables were of Boolean 
type and < was a valid operator on Booleans (False < True, in an ideal 
world) but it's not so (in Pascal).

HTH,
John