Well, you see, this is indeed unusual, but that's the way it is. So to
speak, # is a generalized n-ary operator. So:
a#(b#c)#d
is equivalent to
'#'(1:a 2:b#c 3:d)
I guess you'll just have to live with that :o)
By the way, you can confirm this sort of thing by means of the procedures
Width and Label, as follows:
{Browse {Width a#(b#c)#d}}
{Browse {Label a#(b#c)#d}}
Cheers,
Jorge.
Terrence Brannon escreveu:
> Re:
> http://www.mozart-oz.org/documentation/tutorial/node3.html#chapter.basics
>
> I've been staring at this sentence for 10 minutes, trying to simply
> accept what it says, but it just does not make any sense to me:
>
> observe that 1#2#3 is a single tuple of three elements
>
> But to my way of thinking, # is a binary operator and regardless of
> associativity, it first creates a 2-tuple of two values and then
> another 2-tuple nesting the first 2-tuple and the remaining element.
>
> I really dont see how # could operate any other way, but would
> appreciate any feedback on how this is possible.
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>
Jorge M. Pelizzoni
ICMC - Universidade de São Paulo
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