Optional chaining, but for a variable?

Question

For something like the following:

"use strict";
let a = 1;
console.log(eval(`
    a++; 
    let b=4;
`));
console.log(`a=${a}, b=${b}`);
// ReferenceError: b is not defined

Is it possible to do something where it prints undefined or something if the variable is not there instead of throwing an error? Something conceptually like:

console.log(`a=${a}, b=$?.{b}`)

For standalone identifiers, the only way to see if they exist is to check their `typeof` — CertainPerformance, Mar 07 '22 at 20:02
the problem is not that its undefined, its that its not defined. As in, not declared anywhere — Yftach, Mar 07 '22 at 20:03
@CertainPerformance so maybe this would be the shortest way to do it? ```console.log(`a=${a}, b=${typeof b === 'undefined' ? 'none' : b}`);``` ? — David542, Mar 07 '22 at 20:08

score 0 · Answer 1 · edited Mar 07 '22 at 21:01

0

The placeholder has to contain an expression and there doesn't seem to bean idiomatic way of using an undefined variable (likely due to parsing error) so I would go with a getter instead of using a raw variable.

For example before accessing b you can write

let getB = function () {
    return (typeof b !== "undefined")? b :undefined;
}

Then in the template literal you can write ${getB}.

edited Mar 07 '22 at 21:01

CollinD

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answered Mar 07 '22 at 20:20

uberhaxed

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could you show how that would work or what you mean? If it's a getter, wouldn't you have to pass the variable itself to it which would produce a referenceerror ? – David542 Mar 07 '22 at 20:32
@David542 Getters don't take arguments. In the getter function you can do any type of type checking logic and are able to return a string or a value like null. – uberhaxed Mar 07 '22 at 20:41
ok, do you want to show that in your answer with some code instead of just your single sentence and I can go ahead and accept your answer? – David542 Mar 07 '22 at 20:42

Optional chaining, but for a variable?

1 Answers1