haskell,syntax,infix-notation,applicative,infix-operator

Why is f <$> g <$> x equivalent to (f . g) <$> x ...well, this isn't so much a functor-thing as a Haskell-thing. The reason it works is that functions are functors. Both <$> operators work in different functors! f <$> g is in fact the same as...

Each of these parenthesized expressions but (-2) (edit: and ((-) 2)) are sections, i.e. functions that take one argument and "put it on the missing side of the infix operator" (see this haskell.org wiki). (-2) is not a function, but a number (negative 2): λ> :t (-2) (-2) :: Num...

Python has the operator module, which includes a operator.not_() function: import operator map(operator.not_, (True, False, False)) not itself is one of the boolean operators....

It is not possible to redefine any traits. You could create your own trait with the same name and same methods, which is what you have done. However, the operator + is tied to std::ops::Add, so it wouldn't be useful in this case. In your case, it looks like you...

java,infix-notation,postfix-operator,infix-operator

Spaces You are not putting any spaces on your postfix variable. You are only checking if the current character is one of the "interesting" characters (digits, operators), but not whether it's a space. As a result, if the current character is a space, you just skip it and you don't...

haskell,types,constructor,infix-operator

It's not part of the Haskell standard, but as jamshidh mentions it is still possible in GHC. The caveat is that data constructors (not type constructors) must start with a colon: {-# LANGUAGE TypeOperators #-} data a + b = a :+ b f :: a + b -> a...