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Re: reduce metaoperator on an empty list

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Stuart Cook
May 18, 2005 18:27
Re: reduce metaoperator on an empty list
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On 5/19/05, Matt Fowles <> wrote:
> All~
> What does the reduce metaoperator do with an empty list?

/me puts on his lambda hat

In Haskell, there is a distinction between foldl and foldl1 (similar
remarks apply to foldr/foldr1[1]):

The former (foldl) requires you to give an explicit 'left unit'[2],
which is implicitly added to the left of the list being reduced. This
means that folding an empty list will just give you the unit.

foldl (+) 0 [a,b,c] = ((0+a)+b)+c
foldl (+) 0 [] = 0

The latter (foldl1) doesn't use a unit value, but this means that you
can't fold empty lists.

foldl1 (+) [a,b,c] = (a+b)+c
foldl1 (+) [] = ***error***

/me puts camel hat back on

This suggests that []-reducing an empty list should probably (IMO) do
one of the following:

* Fail, since it's an ill-defined operation without an explicit unit
* Try to find a suitable unit value (see below), and fail if it doesn't find one

You /could/ try to do something 'sensible' and return 0 or undef, but
this seems likely to result in more confusion.

Another alternative is to give the user the option of specifying such
a unit when using the reduction meta-operator, but this seems to work
against the whole point of [+] (which is brevity). If you want to
specify your own unit, use '&reduce'.

> my @a;
> [+] @a; # 0? exception?
> [*] @a; # 1? exception?
> [<] @a; # false?
> [||] @a; # false?
> [&&] @a; # true?
> Also if it magically supplies some correct like the above, how does it
> know what that value is?

Perhaps the operator could have some kind of 'unit' trait? (Or perhaps
'left_unit' and 'right_unit'?)


[1] Just remember that unlike foldr/foldl, which are explicitly
right/left associative, [+] is 'DWIM-associative', reflecting the
associativity of the underlying operator.

[2] e.g. 0 is the left (and right) unit of + because 0 + x == x (and x + 0 == 0)

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