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So? They still have a type, it doesn't have to be "screw you figure it out yourself". Functor f => (a -> b) -> f a -> f b
by asdkjh345fd 6y ago
So? They still have a type, it doesn't have to be "screw you figure it out yourself".
Functor f => (a -> b) -> f a -> f b
- jhoechtl 6y agoRepeating the same over and over again doesn't make it any clearer for the ones trying to follow your line of argumentation. Seems like you had some deep exposure to Haskell, ML or Hindley-Milner in general which, when excessively consumed, detaches from reality. For one reason or the other you take this discussion serious and personal.
- asdkjh345fd 6y agoIf something is unclear, you can ask for clarification. I repeated the statement to three people because three people repeated the same argument to me. This is how conversations work. I have very limited exposure to haskell, am not detached from reality, and am taking nothing personal. Of course the discussion is serious, why would I spend time engaging in a frivolous and meaningless discussion?
- jcelerier 6y agocould you provide the type of this map function ? auto map(auto x) { if constexpr(is_same<decltype(x), int>) { struct T1 { int getStuff() { return 0; } }; return T1 {}; } else { struct T2 { void doStuff() { } }; return T2 {}; } }
- asdkjh345fd 6y agoRead the rest of the discussion.
- welkam 6y agoI have seen few videos by Walter Bright, Andrei Alexandrescu on ranges and I have no problem understanding D`s documentation. Maybe you should learn the language before using it.