What would you expect
foo [id, \x -> x]
to be?
Martin Hofmann wrote on 15.05.2009 12:09:
It is pretty clear, that the following is not a valid Haskell pattern:
foo (x:x:xs) = x:xs
My questions is _why_ this is not allowed. IMHO, the semantics should be
clear: The pattern is expected to succeed, iff 'x' is each time bound to
the same term.
Isn't this allowed, because this would require a strict evaluation of
the 'x' variables?
Thanks,
Martin
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