Improved documentation for Bool (Was: [Haskell-cafe] Comments
from OCaml Hacker Brian Hurt)
daniel.is.fischer at web.de
Sun Jan 18 12:13:21 EST 2009
Am Sonntag, 18. Januar 2009 17:48 schrieb roconnor at theorem.ca:
> 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 at 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.
Thanks. Really helpful. A few minor typos, though.
> -- |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
You have the arrows i and j pointing in the wrong direction.
> -- In Haskell we call we define 'False' to be i(*) and 'True' to be j(*)
Delete "we call".
> -- 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 at .
> -- Similarly from the other commuting triangle we see that
> -- (u+v)(j(*)) = v(*), which means
> -- @if False then a else b = b@
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