[Haskell-cafe] Re: Laws and partial values
daniel.is.fischer at web.de
Sat Jan 24 15:31:30 EST 2009
Am Samstag, 24. Januar 2009 21:12 schrieb Thomas Davie:
> On 24 Jan 2009, at 20:28, Jake McArthur wrote:
> > Thomas Davie wrote:
> >> But, as there is only one value in the Unit type, all values we
> >> have no information about must surely be that value
> > The flaw in your logic is your assumption that the Unit type has
> > only one value. Consider
> > bottom :: ()
> > bottom = undefined
> > Oviously, bottom is not ()
> Why is this obvious – I would argue that it's "obvious" that bottom
> *is* () – the data type definition says there's only one value in the
> type. Any value that I haven't defined yet must be in the set, and
> it's a single element set, so it *must* be ().
It's obvious because () is a defined value, while bottom is not - per
The matter is that besides the elements declared in the datatype definition,
every Haskell type also contains bottom.
> > , but its type, nonetheless, is Unit. Unit actually has both () and
> > _|_. More generally, _|_ inhabits every Haskell type, even types
> > with no constructors (which itself requires a GHC extension, of
> > course):
> Does it? Do you have a document that defines Haskell types that way?
Section 4.2.1 of the report
> > data Empty
> > bottom' :: Empty
> > bottom' = undefined
> > If you only ever use total functions then you can get away with not
> > accounting for _|_. Perhaps ironically a function that doesn't
> > account for _|_ may be viewed philosophically as a partial function
> > since its contract doesn't accommodate all possible values.
> Now that one is interesting, I would argue that this is a potential
> flaw in the type extension – values in the set defined here do not
> exist, that's what the data definition says.
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