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Is ?. a valid operator in Typescript or Javascript?

I have seen code like below:

const var1 = obj1?.innerObj1?.somePropVal;

When I try this in the chrome console, I get an error if obj1 is undefined. Not sure whether this kind of code would work.

Let me know if I am missing something.

Sebastian Simon
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cethie
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    optional chaining (?.) is valid in both TypeScript and relatively recently also vanilla JS. The above code should work in chrome if you're using a somewhat up to date chrome – Nick Parsons Oct 05 '21 at 08:56
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    What do you mean with _"obj1 is undefined"_? I get an error when obj1 is not defined but I don't get an error when it's undefined. `undefined` and "not defined" are two different things in JavaScript. You can't use optional chaining to catch `ReferenceError`. – jabaa Oct 05 '21 at 08:58
  • Are you talking about ternary operator ( ? : ) ? – itiDi Oct 05 '21 at 08:59
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    @itiDi: OP is clearly not talking about that. `?.` is mentioned in the title, in the text and in the code example. It's highly unlikely that they mistyped 3 times and didn't notice. Also if you replace `?.` with `?:` in the sample, it makes absolutely no sense. – Joachim Sauer Oct 05 '21 at 09:01
  • @itiDi I was talking about the optional chaining operator. I got the required explanation from other comments. Thank you. – cethie Oct 05 '21 at 09:05
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    @NickParsons I was able to read up on "optional chaining" and understand. Thanks a ton for helping with the term. That was the key. – cethie Oct 05 '21 at 09:07

1 Answers1

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const var1 = obj1?.innerObj1?.somePropVal;

is the same as:

const var1 = obj1 && obj1.innerObj1 ? obj1.innerObj1.somePropVal : undefined;
Old Geezer
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