16 Apr
2008
16 Apr
'08
10:08 a.m.
On 16 Apr 2008, at 15:14, Miguel Mitrofanov wrote:
Before somebody noticed: I'm wrong.
It's not List monad, but also a "(->) x" monad, also defined in Control.Monad.
Therefore, "return y" is just "const y". Therefore,
x >>= (return y) = x >>= (const y) = x >> y
Right. It is an interesting monad, but it may cause unexpected effect, in view of its implicit name. Hans