[Haskell-cafe] Are all arrows functors?
dave at zednenem.com
Thu Nov 5 16:55:19 EST 2009
On Thu, Nov 5, 2009 at 4:34 PM, Andrew Coppin
<andrewcoppin at btinternet.com> wrote:
> Nicolas Pouillard wrote:
>> Excerpts from Neil Brown's message of Tue Nov 03 13:45:42 +0100 2009:
>>> I was thinking about some of my code today, and I realised that where I
>>> have an arrow in my code, A b c, the type (A b) is also a functor. The
>>> definition is (see
>>> fmap = (^<<)
>>> -- Or, in long form:
>>> fmap f x = arr f <<< x
>>> Out of curiosity, and since this is a typical haskell-cafe question, does
>>> this definition of fmap hold for all arrows?
>> Yes, as shown by the 'WrappedArrow' newtype:
> While I don't wish to suggest that "all arrows are functors" is false, I
> think the argument "yes, because this library says so" is not too strong.
> Let us not forget, according to *the libraries*, Double is in Enum - which,
> I think you'll agree, is just weird...
It's fairly simple to prove the functor laws using the arrow laws.
Among the nine laws for arrows are
a >>> arr id = a
a >>> arr f >>> arr g = a >>> arr (g . f)
Using the definition fmap f a = a >>> arr f, it's pretty simple to
prove the functor laws:
fmap id = id
fmap f . fmap g = fmap (f . g)
Dave Menendez <dave at zednenem.com>
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