roconnor@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 libraries@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