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
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