What type functions mean on a language level

[Jacob Carlborg]

On 2021-01-02 10:28, Stefan Koch wrote:

> The issue here is knowing when the type is complete.
> Imagine you did not put int into there, but some template or another 
> typefunction result.
> At which point can I take the lock and register myself in the type 
> universe?
> And how do I make sure I don't have to type LinkedList(f(x, g(y), j(z))) 
> to reference the type?
> 
> I guess I could create the name to just be "LinkedList(f(x, g(y), 
> j(z)))" where f(x, g(y), j(z))) is eagerly evaluated, and then the name 
> is LinkedList(float) or something like that.
> 
> But that seems like reinventing templates with all their troubles...

I don't know. I'm not the expert.

> perhaps it's still worthwhile though?

I like the syntax at least :)

Your example doesn't have the same problem because it's explicitly 
calling `__internal__magic__registerType`?


> Hmm I wanted to avoid having to puzzle ast nodes together, as I find it 
> rather counter verbose to use.
> simple stuff like;
> mixin(f(a, b));
> becomes
> new CallExp(findFunction("f"), [findVariableExpression("a"), 
> findVariableExpression("b")].insertIntoTreeInPlace();

There's nothing that prevents building helper functions on top of the 
low level AST.

In my idea for AST macros, there would be support for quasi-quoting and 
splicing. It's implemented as a built-in function that converts the 
internal AST (defined in the compiler) to the external AST (defined in 
druntime).

ast(f(a, b)); // single line

ast({ // multiple statements
     int a = 3;
     int b = 4;
     f(a, b);
});

auto name = ast(a);

ast({
     int $name = 3; // splicing. This will declare a variable with the 
name "a"
});

-- 
/Jacob Carlborg