Re: [Haskell-cafe] type inference question

[minh thu]

2009/10/8 Jochem Berndsen <joc...@functor.nl>:
> minh thu wrote:
>> Also, I'd like to know why
>>
>> id id True
>>
>> is permitted but not
>>
>> (\f -> f f True) id
>
> If you want to do this, answer the question "what is the type of (\f ->
> f f True)"?
> You can do this, by the way, using rank-2 types:
>> {-# LANGUAGE Rank2Types, PatternSignatures #-}
>> thisIsAlwaysTrue = (\ (f :: forall a. a -> a) -> f f True) id
>
> Cheers, Jochem

So I learned we should be explicit about f being used polymorphically;
that's what rank-2 types are for. (correct ?)

Thanks,
Thu
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