
Re: Reason for operator precedence
Posted:
Mar 14, 2006 1:09 PM


matt271829news@yahoo.co.uk wrote: >briggs@encompasserve.org wrote: >>matt271829news@yahoo.co.uk writes: >>>Tony wrote: >>>> >>>> Hope this isn't a silly question. >>>> I was wondering what the reason is for having multiple levels >>>> of operator precedence? Phrased another way, why is it that >>>> we don't just evaluate everything from left to right? Having >>>> multiple levels of precedence obviously adds complexity, so I >>>> assume there must be some payback. However, I don't see what it is. >>> >>> As far as addition/subtraction vs multiplication/division is concerned, >>> one reason is to ensure that the distributive property of >>> multiplication works sensibly. For example, we want 3*(4 + 6) = 3*4 + >>> 3*6 = 3*(6 + 4) = 3*6 + 3*4. >> >> Remember that what we're talking about here is merely a notational >> convention. It has nothing whatsoever to do with the distributive >> property of multiplication over addition. > > And, to elaborate a bit more, I venture to disagree and suggest that > the convention *does* have to do with this property. I suggest that the > distributive property of * over + is one of the reasons  possibly the > main reason  why it is "natural" to view multiplication as "tighter" > than addition, and to want to interpret, say, 3*4 + 3*6 as (3*4) + > (3*6) rather than as ((3 * 4) + 3) * 6 or whatever.
Expanding (X+Y)*Z > X*Z+Y*Z is easy and leads to normal forms.
Factoring (X+Y)*Z < X*Z+Y*Z is hard and leads to nonnormal forms.
Therefore X+Y*Z = X+(Y*Z) not (X+Y)*Z (i.e. expanded, not factored).
E.g. compare representational forms for integers and polynomials. It's easy to compute the normal expanded form of polynomials and the radix form of integers (= expanded polynomial in the radix) but its much harder to compute factorizations. Further, poly factorizations needn't be unique if the coef ring isn't a UFD. Performing arithmetic operations in expanded form is easy but is much harder in factored form, e.g. the sum of prime powers p^m + q^n must be completely factored to retain factored form.
Given such a preference for expanded vs. factored representations it is only natural to specify a notation that is more efficient at representing the preferred expanded form.
Bill Dubuque

