[Haskell-beginners] Confused by type constraints

[Guy]

On 04/06/2011 22:49, Brent Yorgey wrote:
> On Fri, Jun 03, 2011 at 05:13:43PM +0300, Guy wrote:
>> On 03/06/2011 16:55, Daniel Schoepe wrote:
>>> On Fri, 03 Jun 2011 16:04:05 +0300, Guy<guytsalmaves-h at yahoo.com>   wrote:
>>>> I have some code resembling
>>>>
>>>> class Foo f where
>>>> 	foo :: a ->   f a
>>>>
>>>> data Bar f a = Foo f =>   Bar {bar :: f a}
>>>>
>>>> instance Foo (Bar f) where
>>>> 	foo a = Bar $ foo a
>>>>
>>>> GHC insists on Foo f =>   in the instance declaration. However, the definition of Bar guarantees that this will always be
>>>> the case. Why do I have to state this explicitly?
>>>
>>> The constraints in the data type declaration only affect the
>>> constructors. It's a known issue, unfortunately there doesn't seem to be
>>> a way around it.
>>
>> It seems to work for simpler examples. For example, this is OK
>>
>> data X x = Show x =>  X x
>>
>> s :: X x ->  String
>> s (X x) = show x
>>
>> I don't have to add a Show x =>  type constraint to s.
>
> In this case, you are *destructing* an X value, and when you do, the
> Show constraint on x becomes available, since the X constructor is
> always paired with a Show constraint.
>
> In your original example, you are *constructing* a Bar value.  You
> have stated that the Bar constructor must be accompanied by a Foo
> constraint, but there is no such constraint in sight!
>
> So the difference is between *destructing* a value (when the
> associated constraint becomes available) and *constructing* one (when
> the required constraint must be provided).
>
> I don't know if I've explained that well.  As others have mentioned,
> the ability to specify constraints on constructors in this way is
> probably going to be removed from the language anyway.

So in my original example, why isn't "instance Foo (Bar f)" destructing Bar - and making the constraint available?