function application vs. infix operator precedence
Geoffrey G Plitt
Wed, 13 Mar 2002 17:15:20 -0500 (EST)
I have made my own infix operator (let's say it's called $) that
takes an int on the left side, an int->int function on the right
side, and returns a function int->int.
this works [and is the functionality i am going for]: (1 $ f) 2
but this doesn't [this is the syntax i want, though]: 1 $ f 2
i guess f is being applied to 2, which i don't want.
so is there a way to make the $ bind with a higher precedence than the