Can Optional Chaining be alternative of 'in' operator in Javascript

Question

I often saw in operator in some library of Javascript. but I think in operator has risk of causing human errors. because we have to write property name as string type.

I think optional chaining could be alternative of in operator and It looks more safe. we can get help of intelligence of IDE during writing code.

what is the benefit of using in keywords instead of Optional Chaining?

optional chaining wont give valid results if your object has properties that are falsy eg: `{zipCode: 0}`, will not run your if-statement if you do `if(address?.zipCode)`. — Nick Parsons, Jul 03 '22 at 05:49
Proper IDEs will find the in-check as well if you do refactorings. — luk2302, Jul 03 '22 at 05:51
your first if checks if the object has the property only so the value doent matter. in the second if it does the console log only if the value is truthy — cmgchess, Jul 03 '22 at 05:52
Oh... I feel embarrassed :( I didn't think about the case which is `address?.zipCode` is falsey. Thanks for your kind comments! — Jangchun Lee, Jul 03 '22 at 14:36

DecPK · Accepted Answer · 2022-07-03T06:01:45.430

AFAIK, in is used to get to know whether the property exists in an object or its prototype or not.

const address = {
  zipCode: 1234,
  country: 'usa',
  city: 'new York',
};

if ('zipCode' in address) {
  console.log('This is address from in checking');
}

and if you are using optional chaining then you are checking the value to be truthy or not

const address = {
    zipCode: 1234,
    country: 'usa',
    city: 'new York',
};

if (address?.zipCode) {
    console.log('This is address from optional chaining');
}

But think of a situation where the value of that property is a falsy value then the in operator will give you the result of property existence. If the value is a falsy value then you have to make additional checks to know whether the property exists or not.

Value can be anything so you have to make necessary checks in optional chaining

const address = {
  zipCode: 0,
  country: 'usa',
  city: 'new York',
};

if ('zipCode' in address) {
  console.log('This is address from in checking');
}

if (address?.zipCode) {  // Won't work
  console.log('This is address from optional chaining');
}

if (address?.zipCode || address?.zipCode === 0) {  // Will work with extra checks
  console.log('This is address from optional chaining');
}

Nick Vu · Answer 2 · 2022-07-03T06:08:37.827

You're comparing 2 different things

In your context, in operator is similar to Object.hasOwnProperty, and its returned value is true/false.

const address = {
   zipCode: 123
}

console.log('zipCode' in address) //true

console.log(address.hasOwnProperty('zipCode')) //true

Optional chaining is to check undefined values optionally

//the short form of `address && address.zipCode`
console.log(address?.zipCode) //false - because no `address` variable

You can consider this case that zipCode property is there but its value is undefined

const address = {
   zipCode: undefined
}

console.log('zipCode' in address) //true

console.log(address?.zipCode) //undefined!

From the above example, you can see that zipCode is in address, but optional chaining does not help you to check it but returns its value instead.

what is the benefit of using in keywords instead of Optional Chaining?

In the conclusion, you should use in when you want to check properties in that object exist or not. In other cases, you can use optional chaining to not only verify property availability but also its values (0, null, and undefined).

Can Optional Chaining be alternative of 'in' operator in Javascript

2 Answers2