fix(q76): 📝 the answer and its explanation (Fully detailed💯)

[SeyyedKhandon]

Fixed the question/answer and also covered the old question.

76. What's the output?

const { firstName: myName } = { firstName: 'Lydia' };

console.log(firstName);

• A: "Lydia"
• B: "myName"
• C: undefined
• D: ReferenceError

Answer

Answer: D

By using destructuring assignment syntax we can unpack values from arrays, or properties from objects, into distinct variables:

const { firstName } = { firstName: 'Lydia' };
// ES5 version:
// var firstName = { firstName: 'Lydia' }.firstName;

console.log(firstName); // "Lydia"

Also, a property can be unpacked from an object and assigned to a variable with a different name than the object property:

const { firstName: myName } = { firstName: 'Lydia' };
// ES5 version:
// var myName = { firstName: 'Lydia' }.firstName;

console.log(myName); // "Lydia"
console.log(firstName); // Uncaught ReferenceError: firstName is not defined

Therefore, firstName does not exist as a variable, thus attempting to access its value will raise a ReferenceError.

Note: Be aware of the global scope properties:

const { name: myName } = { name: 'Lydia' };

console.log(myName); // "lydia"
console.log(name); // "" ---- Browsers e.g. Chrome
console.log(name); // ReferenceError: name is not defined  ----- NodeJS

Whenever Javascript is unable to find a variable within the current scope, it climbs up the Scope chain and searches for it and if it reaches the top-level scope, aka Global scope, and still doesn't find it, it will throw a ReferenceError.

• In Browsers such as Chrome, name is a deprecated global scope property. In this example, the code is running inside global scope and there is no user defined local variable for name, therefore it searches the predefined variables/properties in the global scope which is in case of browsers, it searches through window object and it will extract the window.name value which is equal to an empty string.

• In NodeJS, there is no such property on the global object, thus attempting to access a non-existent variable will raise a ReferenceError.

[ZahraMirzaei]

more clear with a great explanation.