#46Function expressions

~clover commented · 2 years ago
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We will work on the assumption #33 is implemented and a lack of a return type means inference, not void.

A function statement defines a function name and an implementation:

fn hello(a: int) { 1 }

A function expression (lambda) is the same but without a name.

let hello: T = fn(a) { 1 };

Param types may be omitted if the assignment target has a function type.

A function type expression looks like this:

let T = fn (a: int) -> float;

When this is placed before the let hello line, semantic analysis will assign the function into the type T (this function prototype) and actualize the parameters a and the return type into a float.

We will disallow this for now

let hello = fn(a) { 1 }; // what is the type

We will disallow this for now

fn hello(a) { 1 } // what type is `a`

The function statements may appear either module level or block level. The former allowing use-before-declare.

Milestone: 0.2.0.
~clover changed the status to Todo · 2 years ago
~lino-levan commented · 2 years ago
I could see

let T = fn (a: int) -> float;

being

let T = (a: int) -> float;

but I don't feel strongly about it. Potential future bikeshed.
~clover commented · 2 years ago
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Owner
future bikeshed. the reasoning of my syntax proposal is to ensure all functions are very clearly annotated as such, theoretically making code much more readable. but these are types, so the argument isn't as strong.