[Haskell-cafe] Re: Overloading functions based on arguments?

[John A. De Goes]

Take, for example, this function:

f :: [Char] -> Char

f []     = chr 0
f (c:cs) = chr (ord c + ord (f cs))

[] is typed as [Char], even though it could be typed in infinitely  
many other ways. Demonstrating yet again, that the compiler *does* use  
the additional information that it gathers to assist with typing.

Regards,

John A. De Goes
N-BRAIN, Inc.
The Evolution of Collaboration

http://www.n-brain.net    |    877-376-2724 x 101

On Feb 13, 2009, at 6:31 PM, Robert Greayer wrote:

> -- John A. De Goes wrote:
>
>>> Adding information cannot remove a contradiction from the  
>>> information
>>> set available to the compiler.
>
>> But it can and often does, for example, for [] or 4. What's the  
>> type of either expression without more information?
>
> [] :: [a]
>
> 4 :: Num a => a
>
> Do I win something?
>
>
>