[project @ 2003-09-11 14:20:40 by simonpj]
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Allow recursive dictionaries
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In response to various bleatings, here's a lovely fix that involved simply
inverting two lines of code, to allow recursive dictionaries. Here's
the comment. (typecheck/should_run/tc030 tests it)
Note [RECURSIVE DICTIONARIES]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider
data D r = ZeroD | SuccD (r (D r));
instance (Eq (r (D r))) => Eq (D r) where
ZeroD == ZeroD = True
(SuccD a) == (SuccD b) = a == b
_ == _ = False;
equalDC :: D [] -> D [] -> Bool;
equalDC = (==);
We need to prove (Eq (D [])). Here's how we go:
d1 : Eq (D [])
by instance decl, holds if
d2 : Eq [D []]
where d1 = dfEqD d2
by instance decl of Eq, holds if
d3 : D []
where d2 = dfEqList d2
d1 = dfEqD d2
But now we can "tie the knot" to give
d3 = d1
d2 = dfEqList d2
d1 = dfEqD d2
and it'll even run! The trick is to put the thing we are trying to prove
(in this case Eq (D []) into the database before trying to prove its
contributing clauses.