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Generic.takeWhile is not copy-free #330

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@gksato

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@gksato

This issue is closely related to #182, but it is a contract failure, so I guess we need to do something about this.

The description of this issue is simple: the document of the function Data.Vector.Generic.takeWhile says:

O(n) Yield the longest prefix of elements satisfying the predicate without copying.

However, the function's implementation is:

takeWhile :: Vector v a => (a -> Bool) -> v a -> v a
{-# INLINE takeWhile #-}
takeWhile f = unstream . Bundle.takeWhile f . stream

(See it on Hackage, or on GitHub)

The document says the function is copy-free, but it is obvious from the code that it requires copy when:

  1. it is used against a vector actually living in the heap, and
  2. the produced vector can't fuse away (for example, it is used more than once).

Note that Data.Vector.Generic.dropWhile also had this problem, and that we resolved it on PR #327 by making dropWhile actually copy-free.
On PR #327, we made it possible by letting dropWhile be fusible only in the case the argument vector to the function is already an unstreamed stream.

An obvious solution to this problem is to remove the phrase "without copying" from the documentation. It is completely sensible to choose that way.
The problem is more complex than that of dropWhile, and we cannot utilize the method same as the one used in #327.
I guess I've come up with a solution that makes takeWhile copy-free while preserving all required stream fusion, but it is rather global and complicated, and might let bugs sneak in.

I'm being lazy and I couldn't write up everything at once. I'll explain the difficulty of this problem and the proposed solution making takeWhile copy-free in subsequent comments.

See Also: #182 #327(+#141)

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