3 ms·
That don't work since you need to use `partial` on `mul`. I'm a big fan of the way Haskell manages operators and partial application. double :: Num a => a
by mFixman 8y ago
That don't work since you need to use `partial` on `mul`.
I'm a big fan of the way Haskell manages operators and partial application.
double :: Num a => a -> a
double = (2 *)