[Haskell-beginners] Compiler can't deduce Bool as instance of ToField

Tom Davie tom.davie at gmail.com
Thu Aug 15 20:11:28 CEST 2013


On 15 Aug 2013, at 19:49, Bryan Vicknair <bryanvick at gmail.com> wrote:

>>> Bryan Vicknair:
>>> postgresql-simple declares Bool as an instance of the ToField class.  The
>>> compiler can't deduce that given this simple code however:
>>> 
>>>  import Database.PostgreSQL.Simple.ToField (ToField(..))
>>> 
>>>  foo :: (ToField a) => a
>>>  foo = True
>>> 
>>> 
>>> It fails with this error:
>>> 
>>>  Db.hs:64:7:
>>>      Could not deduce (a ~ Bool)
>>>      from the context (ToField a)
>>>        bound by the type signature for foo :: ToField a => a
>>>        at Db.hs:63:8-23
>>>        `a' is a rigid type variable bound by
>>>            the type signature for foo :: ToField a => a at Db.hs:63:8
>>>      In the expression: True
>>>      In an equation for `foo': foo = True
>>>  Failed, modules loaded: none.
>>> 
>>> What am I missing?
>>> 
> 
>> Oliver Charles 
>> You have declared that foo is *any* type that has a ToField instance,
>> allowing the caller of foo to determine the type at their will. However,
>> your implementation of foo is more specific and requires a is actually Bool
>> and nothing else.
>> On 14 Aug 2013 10:24, "Bryan Vicknair" <bryanvick at gmail.com> wrote:
> 
> I see.  I was confused, because the following works:
> 
>  bar :: (Num a) => a
>  bar = 5
> 
> I guess it is because the literal '5' could be an Int or a Float, and the type
> system doesn't know.
> 
> When I saw that the following does not work...::
> 
>  bar :: (Show a) => a
>  bar = False
> 
> ...it made a bit more sense.  It seems that if a literal can be considered part
> of a typeclass in more than one way, you can declare that literal to be of that
> typeclass.  However, if there is only one way for a literal to belong to a
> typeclass, then you can't.

That’s not it here.  What you’re saying in the first case is “no matter what numeric type you want, I can provide it”.  This is true, because the compiler can decide that 5 can indeed be any numeric type.  In the latter case you make the promise “no matter what type you want, as long as you can show it, I can provide it”, but it’s false.  I can say “well okay, give me an Int” and your function can not provide it to me, despite telling me in its type signature that it can.

Tom Davie





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