New-to-Haskell, i.e. elementary, Haskell questions; extended discussions

Sergey Mironov | 2 Oct 2012 09:55

Copying a syntax tree

Hi! I have a syntax tree defined like this:

>
> data A a = A Int Int (B a)
>
> data B a = B String String (C a)
>
> data C a = C Int Int (D a)
>

and so on, all the data are parametrized with a type variable. This variable
is actually used as a field in the very end of a hierarchy:

>
> data D a = D Int a
>

Now I have to write a function which would copy (A Int) to (A String). Is it
possible to do so using TH/syb without writing

copyA (A i1 i2 b) = A i1 i2 (copyB b)
copyB = ...
copyC = ...
..
copyD (D i a) = D i (show a)
?

Could you provide me with a hint?

Thanks,

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Erik Hesselink | 2 Oct 2012 10:26

Re: Copying a syntax tree

You could add {-# LANGUAGE DeriveFunctor #-}, and then add 'deriving
Functor' to all your data types (or you could of course manually
define your functor instances). Then what you want is just 'fmap
show'.

Erik

On Tue, Oct 2, 2012 at 9:55 AM, Sergey Mironov <ierton <at> gmail.com> wrote:
> Hi! I have a syntax tree defined like this:
>
>>
>> data A a = A Int Int (B a)
>>
>> data B a = B String String (C a)
>>
>> data C a = C Int Int (D a)
>>
>
> and so on, all the data are parametrized with a type variable. This variable
> is actually used as a field in the very end of a hierarchy:
>
>>
>> data D a = D Int a
>>
>
> Now I have to write a function which would copy (A Int) to (A String). Is it
> possible to do so using TH/syb without writing
>
> copyA (A i1 i2 b) = A i1 i2 (copyB b)
> copyB = ...

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Sergey Mironov | 2 Oct 2012 12:33

Re: Copying a syntax tree

Yes, thats good solution. Let me complicate things a bit. Lets define
last data, D, in a different way:

> data A a = A Int Int (B a)
> data B a = B String String (C a)
> data C a = C Int Int (D a)
>
> data D a = D Int (F a)

where F is a type function

> type family F t
> data Stage1
> data Stage2
> type instance F Stage1 = Int
> type instance F Stage2 = String

This time D is not a functor, because 'a' can accept only either
Stage1 or Stage2. And what I need is to quickly define copy :: A
Stage1 -> A Stage2

is it possible without writing boilerplate?

2012/10/2 Erik Hesselink <hesselink <at> gmail.com>:
> You could add {-# LANGUAGE DeriveFunctor #-}, and then add 'deriving
> Functor' to all your data types (or you could of course manually
> define your functor instances). Then what you want is just 'fmap
> show'.
>
> Erik

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