It does seem like it should rotate in 1s, but if it does so, what happens when it rotates out something that isn't all 1s into a negative number or all 0s into a positive number?

We'd expect `rotateL n . rotateR n = id = rotateR n . rotateL n` for the finite cases.

but you can't rotate out the digits "up at infinity".

So even this is messy. =/

Ideally we would move rotate/rotateL/rotateR to FiniteBits, but there is a huge upgrade/migration cost associated with that. All user specified instances break in a way that requires CPP and it diverges from the standard further.

On Mon, Sep 29, 2014 at 11:33 AM, David Feuer <david.feuer@gmail.com> wrote:

Problem:

The documentation currently specifies that for non-finite Bits instances, rotate is the same as shift. This seems reasonable for Integer rotateR, but very strange for Integer rotateL. In particular:

rotateL 5 (-1 :: Int) = -1
rotateL 5 (-1 :: Integer) = 16

Based on the concept that Integers represent two's complement out to infinity, rotating a negative integer left should fill with 1s, but it fills with 0s instead.

Solutions:

As stated in the subject, I think one reasonable solution is just to make rotateL fill with 1s for negative non-finite instances, and the other is to move rotations to FiniteBits, as was done with bitSize.

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