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Subject: Re: [boost] [mpl]iter_fold_if Forward Backward rationale?
From: Larry Evans (cppljevans_at_[hidden])
Date: 2009-04-04 03:05:45

On 04/03/09 21:31, David Abrahams wrote:
> on Thu Apr 02 2009, Larry Evans <> wrote:
>> One reason I'm interested is that I can't figure out how reverse_fold
>> would work on a sequence, S, such that begin<S>::type with just a
>> forward iterator. I would have thought that you'd need a
>> bidirectional iterator in order to traverse backwards; yet:
>> shows that it works on lists.
> More later, but remember there's recursion involved. You apply the
> forward function "on the way in" and the reverse one "on the way out" of
> the recursion.
Thanks David.

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"?


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