Typescript: Syntax suggestion: Ignore last void parameters

[thr0w]

class X<T>
{
    f(t: T) { return { a: t }; }
}


var x: X<number>;
var t1 = x.f(5);
t1.a = 5; // Should not error: t1 should have type {a: number}, instead has type {a: T}

var x2: X<void>; 
var t2a = x.f(void 0); // It works but void 0 it's unnecessary
var t2b = x.f(); // SUGGESTION: ignore syntax check of last argument if it's type is void
t2b.a = void 0; // Should not error: t2 should have type {a: void}, instead has type {a: T}

test case: tests/cases/compiler/genericObjectLitReturnType.ts

maybe the solution is the same for #3356

[weswigham]

Would I be correct in saying that the type signature T|void should be equivalent to T?? (Any type which contains the void type is optional as a parameter) This equality is true in the closure compiler type system (but it uses undefined rather than void for the type name).

[thr0w]

Not always @weswigham, just in last arguments of functions.

class X<T, U>
{
    f(t: T|void, u: U) { ... }
}

t: T it's mandatory because is not the last function argument and we can't lose u: U. My suggestion it's only for last argument. However, the rule can be "last void parameters" because, if X<void,void>, f function can be invoked with f() lossless

I don't understand when you say T|undefined (rather than T|void) equality is true in the closure compiler, can you send me a sample?

Thanks

[weswigham]

Oh, sorry, I was (completely) mistaken - The equality's not true for optional function arguments, only optional record type members. :S (Optional function arguments in closure are denoted with an =, while optional members use ?, hence my mixup.)

[RyanCavanaugh]

Chatting with @danquirk, our proposal (roughly) is that at a call expression, we treat the unprovided required arguments as if they were of type void, then check that those are assignable to the parameter types (modulo any). Leave assignability/subtyping of function types unchanged.

[thr0w]

Yes @RyanCavanaugh, just for call expressions. I don't know how works typescript checker then I can't say if it's better don't throw error if last parameter is void and missing or if it's treat the unprovided required arguments as if they were of type void. But it's what you say, just in call expression leaving assignability/subtyping of function types unchanged.

If I can help you just tell what to do.
Thanks.

[RyanCavanaugh]

Approved. Should be useful for Promise<void>.

[jack-williams]

@RyanCavanaugh What is the status on this issue? I saw that @Kingwl was working on a PR that was abandoned (#20642) with some spec issues raised by mhegazy.

Anyway, I have a fix that works for the given examples as well as generic rest / tuple types here 9f6e28b053f94f71ad49438a9b9822bd4b30bb5e if you want to breathe some life into this issue.

@Kingwl : I didn't mean to conflict with your work (I didn't actually see your old PR until I finished my fix). If you want to take my work and finish the PR I would be fine with that.

[Kingwl]

@jack-williams take it easy, I have already close the pr

[FrogTheFrog]

@jack-williams It does not seem to work with constructors. Is this intentional or were constructors just forgotten?

[jack-williams]

@FrogTheFrog Can you provide an example?

[FrogTheFrog]

@jack-williams Here you go:

class Test<T1, T2 = void> {
    constructor(input1: T1, input2: T2) {
        // Do something
    }
}

const test = new Test({});

I'm getting this error with 3.2.0-dev.20181117:

test.ts:7:14 - error TS2554: Expected 2 arguments, but got 1.

7 const test = new Test({});
               ~~~~~~~~~~~~

  test.ts:2:29
    2     constructor(input1: T1, input2: T2) {
                                  ~~~~~~~~~~
    An argument for 'input2' was not provided.


Found 1 error.

[jack-williams]

@FrogTheFrog Thanks for the example. The issue is not to do with constructors; the issue is the generics. See this comment here in an older PR I made.

For instance, this is fine:

class VoidClass {
    constructor(_x: void) {}
}

new VoidClass(); // ok

Basically the special casing for void does not apply to arguments that are instantiated as type void at the point the argument is called. Getting it to work with generics caused too many problems to ultimately be worth it.

Here is a workaround if you _really_ want it:

class Test<T1, T2 = void> {
    constructor(input1: T1, input2: T2) {
        // Do something
    }

    static makeFactory<T1, T2 = void>(): (input1: T1, input2: T2) => Test<T1,T2> {
        return (input1: T1, input2: T2) => new Test(input1, input2);
    }
}

const factory = Test.makeFactory<number>();
const factoryTwo = Test.makeFactory<number, number>();
const testOne = factory(3); // ok
const testTwo = factoryTwo(3); // not ok

[FrogTheFrog]

@jack-williams I'm just nitpicking at this point. Since I mainly use "options" object for public classes, it does not really bother me. I'm using this simple trick fallback to void type when nothing is provided:

interface TestOptions<T1, T2> {
    input1: T1;
    input2?: T2;
}

class Test<T1, T2 = void> {
    constructor(options: TestOptions<T1, T2>) {
        if (options.input2 !== void 0) {
            // Do something with input2
        }
    }
}

const test = new Test({ input1: {} }); //=> test has Test<{}, void> type

I'm just glad that the void 0 nonsense is finally over thanks to you :D

[garkin]

Why does this fail? Playground. npx tsc --version: Version 3.5.1

function qwe<T = void>(a: T) {}
qwe(); // ERROR: Expected 1 arguments, but got 0. function qwe<void>(a: void): void

function abc(a: void) {}
abc(); // Passes

[jack-williams]

@garkin

Basically the special casing for void does not apply to arguments that are instantiated as type void at the point the argument is called. Getting it to work with generics caused too many problems to ultimately be worth it.

The checks are done before T is replaced with void. This is tracked here: #29131