Purpose

This article is intended to convey my confusion and minor frustration with trying to understand part of Haskell's syntax as I currently learn Haskell.

Since Haskell has many unfamilar concepts relative to other languages I know well, I am likely to make some terminology mistakes here. I am learning and welcome corrections. Please see the discussion on Reddit.

Consider the following function-call expression in Haskell:

```
(alpha beta gamma)
```

What can be deduced?

`alpha`

is a function (either imported or in a variable).

In most languages (that disallow implicit partial application) you could additionally deduce:

`alpha`

takes exactly 2 arguments- The expression’s return type matches that of
`alpha`

.

Neither of those two properties are necessarily true in Haskell.

In this case, the expression behaves as you’d expect in any other language. In particular:

`alpha`

takes exactly 2 arguments- The expression’s return type matches that of
`alpha`

.

In this case, the expression is a partial application of `alpha`

, and would be perhaps better read as:

`(\ ... -> alpha beta gamma ...)`

Or, in a more C-like notation:

`function (...) { return alpha(beta, gamma, ...); }`

Thus:

`alpha`

takes exactly 2+K arguments, for some unknown K.- The expression returns a function of K arguments.

Let’s say `alpha`

takes 1 argument. This would be intuitively read as:

`((alpha beta) gamma)`

After this reformulation you need to recursively examine the new expression.

In this example, we can deduce:

`(alpha beta)`

returns an anonymous function of (at least) 1 argument.- If this anonymous function takes 1 argument, then it is just called with
`gamma`

, and the original expression’s return type matches the return type of the anonymous function. - If this anonymous function takes more than one argument (1+K), then it is partially applied to
`gamma`

, and yet another anonymous function (that takes K parameters) is returned as the result of the original expression.

What a mess. If `alpha`

does not in fact take two arguments then I have to exert non-trivial effort to derive the type of the expression - or even what the expression semantics are.

There are two ways I can see to simplify these weird cases:

Then we would see syntax like:

-- alpha takes 2 arguments -- return type matches that of alpha (alpha beta gamma) -- alpha takes 2+K arguments; -- return type is a partially applied K-argument function (alpha beta gamma ...) -- alpha takes 1 argument and returns a 1-argument function; -- expression's return type matches that of the 1-argument function ((alpha beta) gamma) -- alpha takes 1 argument and returns a (1+K)-argument function; -- expression's return type is a partially applied K-argument function ((alpha beta) gamma ...)

Notice that each conceptually different case now has a unique syntactic representation - it’s no longer just `(alpha beta gamma)`

for all cases.

For example parentheses could be used instead of a space to signify function calls.

Using a grouping operator syntactically prohibits relying on the left-associativity of space for partial function application, since a grouping operator doesn’t have associativity.

If you combined this suggestion with the explicit syntax extension above, you would get syntax like:

-- alpha takes 2 arguments -- return type matches that of alpha alpha(beta, gamma) -- alpha takes 2+K arguments; -- return type is a partially applied K-argument function alpha(beta, gamma, ...) -- alpha takes 1 argument and returns a 1-argument function; -- expression's return type matches that of the 1-argument function alpha(beta)(gamma) -- alpha takes 1 argument and returns a (1+K)-argument function; -- expression's return type is a partially applied K-argument function alpha(beta)(gamma, ...)

- Unique Features of Various Programming Languages
*Describes several other programming languages and their unique features.*

- Visual Guide to Programming Language Properties
*Visualizes how various programming language properties interact.*

I originally got bitten by this syntactic ambiguity when trying to decipher the meaning of:

```
(flip (/) 20)
```

I was not previously familar with `flip`

and so I assumed that it took two arguments, a function (namely `(/)`

) and a non-function (namely `20`

). And that it probably returned a function, since its surrounding context expected a function.

In fact `flip`

takes only one argument and thus is more clearly written as:

```
((flip (/)) 20)
```

And inlined further to be the lower-order:

```
(\x -> x / 20)
```

And, if desired, further rewritten to be the more-compact:

```
(/ 20)
```

This last form uses partial application to fill in the first argument of `/`

, which is obvious since `/`

is a well-known built-in infix operator that requires an unspecified left argument.