Warp wrote:

>   Which got me thinking: Does foldl1 in haskell work for data containers
> other than lists?

No. The function specified in the language standard works only for lists.

However, there *is* an alternate version in one of the libraries that 
does work for any suitable container. (Confusingly, it has the same 
name, even though it's a different function.) Almost all container 
libraries supply their own home-grown fold function, which may or may 
not have the same name and/or type signature.

This is something of a known problem. A container like lists can hold 
*any* type of data, but the Set container can only hold "ordered" data. 
We have yet to reach a concensus about how to encode this into the type 
system. (Dependent types look like the future, but that's an 
experimental new feature that isn't part of any official standard, etc.)

> What if you want to apply it to only a sub-range inside
> the data container rather than to the entire container? (Can you do it
> without copying the sub-range to a separate container?)

Notionally, no.

Physically? Ask the optimiser.

(Remember, your "container" might not actually *exist* in its entirity 
until you try to "do" something with it...)

-- 
http://blog.orphi.me.uk/
http://www.zazzle.com/MathematicalOrchid*

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