Return type of 'super()' should be 'this', not 'void'

[rbuckton]

Found when reviewing #29374. We currently treat the return type of super() as void, but it really should be this.

TypeScript/tests/baselines/reference/derivedClassSuperProperties.errors.txt

Lines 534 to 536 in eb7388a

	for(a; super();) {}
	~~~~~~~
	!!! error TS1345: An expression of type 'void' cannot be tested for truthiness.

This error is interesting because the result of super() is certainly not void. This isn't something you need to address, however.

Originally posted by @rbuckton in #29374 (comment)

[craigphicks]

Why should/must it be 'this'? I am not questioning that it could be this, as a matter of design choice. Does the current return type void cause any problems?

[rbuckton]

Because that's the behavior of super() in a constructor in ES2015. When you call super(), it sets the this value:

class Base {}
class Derived extends Base {
  constructor() {
    let x = super();
    console.log(x === this);
  }
}
new Derived(); // logs: true

[MartinJohns]

Maybe related? #38519

[MartinJohns]

Duplicate of #37847.

[craigphicks]

Because that's the behavior of super() in a constructor in ES2015. When you call super(), it sets the this value:

class Base {}
class Derived extends Base {
  constructor() {
    let x = super();
    console.log(x === this);
  }
}
new Derived(); // logs: true

So the reason 'for it' would be so that JS code could pass the TS compiler.
Are there are examples where JS doesn't pass the compiler? I seem to remember so.

I can see reasons 'against it': x would be a duplicate of this and the compiler has to track that semantically. It's more confusing for a human reader to see two symbols referencing the same object.

[ExE-Boss]

x would be a duplicate of this and the compiler has to track that semantically.

That’s already supported:

let x: this;

[LongTengDao]

Why should/must it be 'this'? I am not questioning that it could be this, as a matter of design choice. Does the current return type void cause any problems?

class MyArray extends Array {
    constructor () { return super(); }
}

It's faster than default constructor.

[ExE-Boss]

For Array, you might want to do:

class MyArray extends Array {
	constructor(length = 0) { return super(+length) }
}

to correctly support ArraySpeciesCreate with non‑0 length.

[typescript-bot]

This issue has been marked as a 'Duplicate' and has seen no recent activity. It has been automatically closed for house-keeping purposes.