## 疑問

> (elem [1,2,3]) 3
True


3 elem [1,2,3]elem 3 [1,2,3]の形ならわかるし，(3 elem) [1,2,3]でも部分適応をしてるのはわかる．しかし，問題のコードがこれでいける理由が謎だった．

## 答え

Stackoverflowに同じような疑問を聞いてる人がいて，その回答を見てわかった．

The Haskell grammar has special support for construct like this, called “operator sections”. If you have any infix operator, like say #$%, then the following notation is supported: (#$%)   = \x y -> x #$% y (#$% y) = \x   -> x #$% y (x #$%) = \y   -> x #\$% y


So you are expecting some mathematical consistency to break this, and if Haskell were a miniscule language like Forth, I would be inclined to agree with your intuition. The reason it works is basically “because they wrote it to work like that”.

この答えを見る限り，やはり (elem [1,2,3])(\ x -> x elem [1,2,3])と等価であるらしい．wiki.haskell.orgのページを見てもやはり同じことが書いてあった．

In Haskell there is a special syntax for partial application on infix operators. Essentially, you only give one of the arguments to the infix operator, and it represents a function which intuitively takes an argument and puts it on the “missing” side of the infix operator.

(2^) (left section) is equivalent to (^) 2, or more verbosely \x -> 2 ^ x
(^2) (right section) is equivalent to flip (^) 2, or more verbosely \x -> x ^ 2


つまり，中置記法の関数に関してはこのような感じで定義がされているということが予想できる．

(infix_op e) = \ x -> x infix_op e
(e infix_op) = \ x -> e infix_op x