According to compose's type at https://wiki.haskell.org/Compose, it can be written as

`compose :: [a -> a] -> (a -> a)`

or `compose :: [a -> a] -> a -> a`

I think these two types is different: the former takes a list of functions and returns a function,the latter takes a list of functions and an argument,then finally returns a value.

Namely, when a function(higher-order function) takes another function as an argument or returns a function as a result, the parentheses around the argument(the result) shouldn't be omitted, e.g., if `filter :: (a -> Bool) -> [a] -> [a]`

removes the parentheses, its meaning will change.

Am I right or wrong?

Answer:

These are both the same:

```
compose1 :: [a -> a] -> (a -> a)
compose1 [] = \x -> x
compose1 [f] = \x -> f x
compose1 (f1:f2:fs) = compose1 ((f2 . f1):fs)
compose2 :: [a -> a] -> a -> a
compose2 [] x = x
compose2 [f] x = f x
compose2 (f1:f2:fs) x = compose2 ((f2 . f1):fs) x
```

Notice how these definitions are virtually identical, except the lambda gets moved from one side of the `=`

to the other. In fact, you can always do the following transformations:

```
f x y z = <expr x y z>
f x y = \z -> <expr x y z>
f x = \y -> \z -> <expr x y z> = \y z -> <expr x y z>
f = \x -> \y -> \z -> <expr x y z> = \x y z -> <expr x y z>
```

This is actually what the compiler does to all your functions. If you compile with `-ddump-simpl`

you will see the Core code dumped where all functions are defined in terms of lambdas. This is because Haskell uses the law that

```
f x = <expr>
```

Is equivalent to

```
f = \x -> <expr>
```

The lambda syntax could be considered a more basic syntax than defining a function using explicit arguments.

Namely, when a function(higher-order function) takes another function as an argument or returns a function as a result, the parentheses around the argument(the result) shouldn't be omitted, e.g., if

`filter :: (a -> Bool) -> [a] -> [a]`

removes the parentheses, its meaning will change.

You are correct in thinking that you can't remove the parentheses from `filter`

's type signature, and this is because function application is right associative only. This means that the following signatures are equivalent:

```
f :: a -> b -> c -> d -> e
f :: a -> (b -> (c -> (d -> e)))
```

This is what is meant by associating to the right, you go from right to left when adding nested parentheses. However, `f`

's signature is **not** equivalent to

```
-- Not equivalent!
f :: (((a -> b) -> c) -> d) -> e
```

Simply because `->`

is not a fully associative operator. For example, with `+`

you have

```
x + (y + z) = (x + y) + z
```

But with `->`

you have

```
x -> (y -> z) /= (x -> y) -> z
```

This is similar to the `:`

operator, e.g.

```
1:2:3:4:[] == 1:(2:(3:(4:[])))
/= (((1:2):3):4):[]
```

The last expression would not type check since `1:2`

is ill typed, `2`

is not a list!

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