Allow instantion expression with empty arguments

🔍 Search Terms

instantiation expression, empty, type arguments

✅ Viability Checklist

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This isn't a request to add a new utility type: https://github.com/microsoft/TypeScript/wiki/No-New-Utility-Types
• This feature would agree with the rest of our Design Goals: https://github.com/Microsoft/TypeScript/wiki/TypeScript-Design-Goals

⭐ Suggestion

I would like to be able to instantiate a generic function type with an empty type arguments list.

It would be nice to allow this for generic types, but it's not as necessary, since a generic type is always instantiated in a type context, even if no type arguments list is given. A generic function type needs to be specifically instantiated.

📃 Motivating Example

Here is a generic function where a type parameter has a default. I can get use an instantiation expression to specialize it to a non-generic function.

declare function find1<T, S extends T=T>(Ar:T[], predicate:(t:T)=>t is S): S | undefined;
type Find1Object = typeof find1<object>;

Here is a generic function with only one type parameter. I cannot use an instantiation expression with the default for S:

declare function find2<S extends object = object>(Ar:object[], predicate:(t:object)=>t is S) : S | undefined;
type Find2 = typeof find2<>; // <- this is a syntax error "Type argument list cannot be empty"

💻 Use Cases

1. What do you want to use this for?
2. What shortcomings exist with current approaches?
3. What workarounds are you using in the meantime?
   • Declare an unused "dummy" type parameter so there's always a non-defaulted type argument.

Originally mentioned here: https://github.com/microsoft/TypeScript/issues/57463#issuecomment-1957739330

L’issue non indica file né test; inizia individuando i percorsi del parser e del controllo dei tipi per le espressioni di istanziazione e la diagnostica che rifiuta un elenco vuoto di argomenti di tipo. Usa l’esempio find2 come caso di accettazione: un elenco vuoto dovrebbe essere accettato e specializzare la funzione generica usando i suoi valori predefiniti, senza modificare il JavaScript emesso.