Subject: Re: [boost] [mpl]iter_fold_if Forward Backward rationale?
From: Larry Evans (cppljevans_at_[hidden])
Date: 2009-04-04 13:41:50
On 04/04/09 02:05, Larry Evans wrote:
> On 04/03/09 21:31, David Abrahams wrote:
>> You apply the forward function "on the way in" and the reverse one
>> "on the way out" of the recursion.
> I thought it would be interesting to compare the mpl folds with the
> haskell folds.
> From pp. 116-117 of:
> the following is a trace of the execution of foldl
> and foldr where the argument order is modified to reflect that of mpl
> fold argument order. IOW:
> [x1,x2,x3] represents an mpl sequence
> z represents the initial state, e.g. list<>.
> F represents the binary operation, e.g.push_front.
> Now, the foldl trace:
> = foldl([x2,x3],F(z,x1),F)
> = foldl([x3],F(F(z,x1),x2),F)
> = F(F(F(z,x1),x2),x3)
> which looks like mpl::fold since, when F == push_front, the resulting
> sequence is reversed, which is what happens with mpl::fold.
> Then, the foldr trace:
> = F(x1,foldr([x2,x3],z,F))
> = F(x1,F(x2,foldr([x3],z,F)))
> = F(x1,F(x2,F(x3,z)))
> which looks like mpl::fold_reverse since, when F == push_front, the
> resulting sequence is the same is the start sequence, which is what
> happens with mpl::fold_reverse. OOPS, push_front takes the sequence
> as 1st arg, but F(x3,z) has the sequence (e.g. list<> ) as the 2nd
> Now, is the F in the foldr applied "on the way out" and the F in the
> foldl applied " on the way in"?
So now I'm wondering what's the haskell counterpart to iter_fold_if.
Assuming ForwardPredicate and BackwardPredicate are always<true_>,
then AFAICT, it would be foldlr defined partially by cases as:
Where F is the iter_fold_if ForwardOp and B is the iter_fold_if
Now, if B==F==push_front, this would produce a palindrome:
This is demonstrated by the attached program which produces on cout:
However, I 've not figured out where the extra number (1 in case of
iter_fold_if<1> and 2 in case of iter_fold_if<2>) is coming from.
OTOH, I haven't tried very hard to figure it out.
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