[Haskell-cafe] Word rigid in "`a' is a rigid type variable..."
Vlatko Basic
vlatko.basic at gmail.com
Thu Nov 14 11:14:41 UTC 2013
Hi Daniel,
Yes, I do understand all that. :-)
The function is the most trivial example to show my confusion with the word
rigid, not the error itself.
I had a feeling Brandon was talking about some important differences in the way
of thinking.
vlatko
-------- Original Message --------
Subject: Re: [Haskell-cafe] Word rigid in "`a' is a rigid type variable..."
From: Daniel Trstenjak <daniel.trstenjak at gmail.com>
To: haskell-cafe at haskell.org
Date: 14.11.2013 12:08
>
> Hi Vlatko,
>
> On Thu, Nov 14, 2013 at 11:33:15AM +0100, Vlatko Basic wrote:
>> Yes, sometimes I still have the feeling I'm not thinking fully Haskellish.
>> Maybe it's time form me to re-read about the type system.
>>
>> Can you recommend any resources that helped you in better understanding?
>
> Did you understand what Brandon said, that 'a' has to be a 'String',
> because it's compared by an other string with '==' and the operator
> '==' demands, that both arguments have the same type?
>
>
> If you try to write the same function in C++ (the same may hold for Java Generics):
>
> template <typename A>
> bool f(A a) { return a == "x"; }
>
>
> Than the C++ compiler would happily read this function and wouldn't
> complain in any way. Only if you would use this function with a type
> that isn't comparable with a string, than it would complain.
>
>
> The problem with the C++ way is, that the function isn't telling you in
> any way the constraints of type 'A', only if you're using a type that
> doesn't fulfill all constraints, than the compiler will tell you that with
> some very indirect error messages.
>
>
> Haskell is very explicit about the constraints, you have to explicitly
> declare all the constraints of your type.
>
>
> So looking at the function 'f':
>
> f :: a -> Bool
> f a = let b = "x" in a == b
>
>
> The first thing is the usage of '==', which is an operator of the
> type class 'Eq', so only types having an instance of 'Eq' are allowed:
>
> f :: Eq a => a -> Bool
> f a = let b = "x" in a == b
>
>
> But you're comparing the 'a' against a variable of type 'String' and
> because the type of '==' is 'a -> a -> Bool' the type of 'a' also has
> to be 'String'.
>
> f :: String -> Bool
> f a = let b = "x" in a == b
>
>
> Greetings,
> Daniel
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