|On a similar note, what does this (legal Haskell 98) definition mean?
|
| (((+) + 1) + 1) 1 = (+)
|
|That's right!
|
| (+) :: (Integral a, Num b, Num c) => a -> b -> c -> a
I was about to ask you to explain this, but in writing this email,
I've worked it out.. :-)
decl -> gendecl
| (funlhs | pat0) rhs
funlhs -> var apat {apat }
| pati+1 varop(a,i) pati+1
| lpati varop(l,i) pati+1
| pati+1 varop(r,i) rpati
| ( funlhs ) apat {apat }
We have: "(((+) + 1) + 1)" then "1", so we can't have <pat0>, so it
must be <funlhs>. There are only two terms, and the first is not
<var> so it must be <(funlhs) apat>. For "((+) + 1) + 1" to be
<funlhs>, the third "+" must be a <varop> as it cannot be an <apat>
(it would have to have brackets around it). So, "((+) + 1)" is a
<pat>, and thus "(+)" is <var>. Therefore, "(+)" is acting as an
integer here.
But then, to what does the "(+)" on the rhs refer? This is similar
to:
f f = f
which was on the list a while back IIRC, and defines the identity
according to Hugs (I'm not sure if this is dictated by the language
report or not). By analogy then, the equation:
(((+) + 1) + 1) 1 = (+)
defines a function (+) that takes an integer N, and two further
elements of Num, and if these elements are both 1 then it returns N-1.
Correct? :-)
There, that only took about an hour... ;-)
Graeme.
