rocon...@theorem.ca wrote:
On Sun, 18 Jan 2009, Ross Paterson wrote:
Anyone can check out the darcs repos for the libraries, and post
suggested improvements to the documentation to librar...@haskell.org
(though you have to subscribe). It doesn't even have to be a patch.
Sure, it could be smoother, but there's hardly a flood of contributions.
I noticed the Bool datatype isn't well documented. Since Bool is not
a common English word, I figured it could use some haddock to help
clarify it for newcomers.
-- |The Bool datatype is named after George Boole (1815-1864).
-- The Bool type is the coproduct of the terminal object with itself.
-- As a coproduct, it comes with two maps i : 1 -> 1+1 and j : 1 -> 1+1
-- such that for any Y and maps u: 1 -> Y and v: 1 -> Y, there is a
unique -- map (u+v): 1+1 -> Y such that (u+v) . i = u, and (u+v) . j = v
-- as shown in the diagram below.
--
-- 1 -- u --> Y
-- ^ ^^
-- | / |
-- i u + v v
-- | / |
-- 1+1 - j --> 1
--
-- In Haskell we call we define 'False' to be i(*) and 'True' to be
j(*) -- where *:1.
-- Furthermore, if Y is any type, and we are given a:Y and b:Y, then
we -- can define u(*) = a and v(*) = b.
-- From the above there is a unique map (u + v) : 1+1 -> Y,
-- or in other words, (u+v) : Bool -> Y.
-- Haskell has a built in syntax for this map:
-- @if z then a else b@ equals (u+v)(z).
--
-- From the commuting triangle in the diagram we see that
-- (u+v)(i(*)) = u(*).
-- Translated into Haskell notation, this law reads
-- @if True then a else b = a...@.
-- Similarly from the other commuting triangle we see that
-- (u+v)(j(*)) = v(*), which means
-- @if False then a else b = b@
I'm going to go ahead and assume this was a joke and crack up... The sad
part is I didn't actually find this difficult to read...
Cory "lost touch with the real world" Knapp
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