
Re: Reason for operator precedence
Posted:
Mar 14, 2006 8:55 AM


bri...@encompasserve.org wrote: > In article <1142342196.542632.294210@i39g2000cwa.googlegroups.com>, matt271829news@yahoo.co.uk writes: > > > > Tony wrote: > >> Hi all. > >> > >> 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. > > You can express the distributive law for multiplication over division > using parentheses: > > a*(b+c) = (a*b) + (b*c)
Obviously you can. I meant to make it work without needing parentheses, but it seems that wasn't clear.

