On 5/19/05, Matt Fowles <[EMAIL PROTECTED]> 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. i.e. 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. i.e. 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'?) Stuart [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)
