You should read 'non negative' where I wrote 'nonzero'.

R.E. Boss


> -----Oorspronkelijk bericht-----
> Van: [email protected] [mailto:programming-
> [email protected]] Namens R.E. Boss
> Verzonden: zondag 1 augustus 2010 12:46
> Aan: 'Programming forum'
> Onderwerp: Re: [Jprogramming] Order of operations
>
> Perhaps it has something to do with the following.
> Multiplication (by a nonzero integer) is repeated addition and
> exponentiation (by a nonzero integer) is repeated multiplication.
> So it is more natural to expand * first and interpret a*b+c as a+...+a +b
> where there are b a's.
> Same with ^ and * .
>
> If a↑↑b is interpreted as a↑....↑a = a^....^a with b a's, cf.
http://en.
> wikipedia.org/wiki/Knuth's_up-arrow_notation than that would have
> precedence
> over ^ as well.
> In general, more up-arrows have precedence over fewer for the very same
> reason: they need to be expanded first.
>
> NB. ↑ is right associative, ie. a↑b↑c = a↑(b↑c) and is not equal to
(a
> ↑b)↑c, which seems natural for J users.
>
> (I you really want a mind dazzling notation, look at
> http://en.wikipedia.org/wiki/Conway_chained_arrow_notation )
>
>
> R.E. Boss
>
>
> > -----Oorspronkelijk bericht-----
> > Van: [email protected] [mailto:programming-
> > [email protected]] Namens Dan Bron
> > Verzonden: zaterdag 31 juli 2010 17:22
> > Aan: 'Programming forum'
> > Onderwerp: [Jprogramming] Order of operations
> >
> > How did the current mathematical order of operations (PEMDAS) arise?
> >
> >
> >
> > Why does it always seem that the more powerful operators also have the
> > higher precedence?  Is it so that expressions don't explode by default
> > (e.g.
> > 3*4^5*6 would have a very different value if * was higher precedence
> than
> > ^).  If we introduce a new operator in the chain + * ^, would it
> > necessarily
> > also have the highest precedence?
> >
> >
> >
> > -Dan
> >
> > ----------------------------------------------------------------------
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>
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