An error occurs in typescript when Optional Chaining is specified as a variable

Question

I encountered the following error when I wrote code using 'Optional Changing'.

type Foo = {
  b?: number
}

const foo:Foo = {}

if (foo?.b) {
  console.log(foo.b / 10)
}

type Foo = {
  b?: number
}

const foo:Foo = {}

const flag = foo?.b

if (flag) {
  // ERROR: Object is possibly 'undefined'
  console.log(foo.b / 10)
}

The first code is no problem.

However, the second code will cause an error.

Why does this error occur?

Also, I would like to know how to avoid the error by writing the second code!

if(foo?.b) {...} would work, but that's not the way you want it to work, right? — Achim
– Achim, Commented Jan 17, 2021 at 8:37
flag is just a variable for typescript, it 'forgets' that it is initialized from foo?.b — D Pro
– D Pro, Commented Jan 17, 2021 at 8:41
@ddd you can use foo.b! / 10 to inform ts that foo.b will be defined at that time — Ramesh Reddy
– Ramesh Reddy, Commented Jan 17, 2021 at 8:46
@D Pro I understand that type inference is not working in the second way. But is there any way to make it work? — ddd
– ddd, Commented Jan 17, 2021 at 8:47

Nishant · Accepted Answer · 2021-01-17 09:18:35Z

3

Typescript does not track related variables as stated in this issue: https://github.com/microsoft/TypeScript/issues/30539. This ticket also has a link to a larger discussion on how control flow analysis works in TypeScript. Put simply, in the if block, the foo.b is not the same as flag.

What you may do here, instead:

type Foo = {
  b?: number
}

const foo:Foo = {}

const flag = foo?.b

if (flag) {
  console.log(flag / 10)
}

answered Jan 17, 2021 at 9:18

Nishant

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