@andy_snow @damncabbage not a functor :(
@gregmcintyre @mwotton @damncabbage Functor is (x->y)->(Fx->Fy) satisfying id and comp. No such thing here, even approximately (like C++).
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@dibblego@mwotton@damncabbage In terms of the OT, I didn't mind Rob's amended code (2nd tweet). Not as nice as Haskell but easy to grok. -
@gregmcintyre@mwotton@damncabbage I think I missed that one. Twitter is awesome like that. - Show replies
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