Union type inference lost in callback in else if clause

[rhys-vdw]

TypeScript Version: 2.9.0-dev.20180418

Search Terms:

infer callback lost null union

Code

My attempt at a minimal reproduction - someone could probably do better:

function takeString(s: string): string {
  return s
}

const x = {
   foo: Math.random() > 0.5 ? "hello" : null
}
if (x.foo === null) {
  console.log("bar")
} else if (["hello"].some(e => e === takeString(x.foo))) { // <-- error here
  console.log("foo")
}

Expected behavior:

No error

Actual behavior:

Argument of type 'string | null' is not assignable to parameter of type 'string'.
Type 'null' is not assignable to type 'string'.

Playground Link: link

Related Issues:

[MartinJohns]

It is working as intended, and the issue is the expression/lambda, not the else-if. The compiler has no guarantee that .some() will call your lambda immediately, so theoretically the value of x.foo could change in the meantime.

A solution would be to have an else block, then assign the correctly typed x.foo to a local const variable and pass that to the function.

#7719 (comment)

[rhys-vdw]

Ah of course, thanks!

Although I should point out that this also fails when the field is readonly. So perhaps there is room for improvement here?

[MartinJohns]

The readonly modifier is not part of the actual type of the foo property, unfortunately. The readonly modifier applies to the interface and declares that the property foo is read-only. As an unfortunate result you can just pass a readonly interface to a method accepting a mutable version and modify the object:

interface ReadonlyTest { readonly foo: string; }
interface MutableTest { foo: string; }

function mutateTest(test: MutableTest): void {
    test.foo = 'oh no!';
}

const readonlyTest: ReadonlyTest = { foo: 'bar' };
mutateTest(readonlyTest);

[RyanCavanaugh]

See #11498

[typescript-bot]

Automatically closing this issue for housekeeping purposes. The issue labels indicate that it is unactionable at the moment or has already been addressed.