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ES 6 Interview Question and Answer - Intermediate Level

The interview questions posted in the article are based on let, rest/spread, default parameter, destructuring, object/template literals, etc introduced in ES 6.
Find the output of the code in each question.

  // Question 1 let a = 2; if (a > 1) { let b = a * 3; console.log(b); for (let i = a; i <= b; i++) { let j = i + 10; console.log(j); } let c = a + b; console.log(c); }

  // Question 2 function foo() { console.log(a); console.log(b);
var a; let b; } foo();

  //Question 3 var funcs = []; for (let i = 0; i < 5; i++) { funcs.push(function() { console.log(i); }); } funcs[3]();

//Question 4 var a = [2, 3, 4]; var b = [1, ...a, 5]; console.log(b);


//Question 5
function foo(x, y, ...z) { console.log(x, y, z); } foo(1, 2, 3, 4, 5);

//Question 6
  function foo(x = 11, y = 31) { console.log(x + y); } foo(); foo(5, 6); foo(0, 42);
foo(5); foo(5, undefined); foo(5, null); foo(undefined, 6); foo(null, 6);

//Question 7
function bar(val) { console.log("bar called!"); return y + val; } function foo(x = y + 3, z = bar(x)) { console.log(x, z); } var y = 5; foo(); foo(10); y = 6; foo(undefined, 10);

//Question 8
var w = 1, z = 2; function foo(x = w + 1, y = x + 1, z = z + 1) { console.log(x, y, z); } foo();

//Question 9
function foo() { return [1, 2, 3]; } function bar() { return { x: 4, y: 5, z: 6 }; } var [a, b, c] = foo(); var { x, y, z } = bar(); console.log(a, b, c); console.log(x, y, z);

//Question 10
var { x, y, z } = bar(); console.log( x, y, z ); var [,b] = foo(); var { x, z } = bar(); console.log( b, x, z );

//Question 11
function foo( [ x, y ] ) { console.log( x, y ); } foo( [ 1, 2 ] ); foo( [ 1 ] ); foo( [] ); function baz( { x, y } ) { console.log( x, y ); } baz( { y: 1, x: 2 } ); baz( { y: 42 } ); baz( {} );
Lets checkout the score. Match your solution with following answers. These question are taken from examples discussed in book You Don't know JS.

Output 1:
// 6
// 12 13 14 15 16
// 8
Reason: Variables declared with let have block scope.

Output 2:
// undefined
// ReferenceError!
Reason: Variables declared with let doesn't gets hoisted.

Output 3:
// 3
Reason: The let i in the for header declares an i not just for the for loop itself, but it redeclares a new i for each iteration of the loop. That means that closures created inside the loop iteration close over those per-iteration variables the way you'd expect.
If you tried that same snippet but with var i in the for loop header, you'd get 5 instead of 3, because there'd only be one i in the outer scope that was closed over, instead of a new i for each iteration's function to close over.

Output 4:
// [1,2,3,4,5]
Reason: In this usage, ... is basically replacing concat(..), as it behaves like [1].concat( a, [5] )here.

Output 5:
// 1 2 [3,4,5]
Reason: The ...z in this snippet is essentially saying: "gather the rest of the arguments (if any) into an array called z." Because x was assigned 1, and y was assigned 2, the rest of the arguments 3, 4, and 5 were gathered into z. 

Output 6:
foo(); // 42 foo(5, 6); // 11 foo(0, 42); // 42 foo(5); // 36 foo(5, undefined); // 36 <-- `undefined` is missing foo(5, null); // 5 <-- null coerces to `0` foo(undefined, 6); // 17 <-- `undefined` is missing foo(null, 6); // 6 <-- null coerces to `0`

Output 7:
foo(); // "bar called" // 8 13 foo(10); // "bar called" // 10 15 y = 6; foo(undefined, 10); // 9 10

Output 8:
foo(); // ReferenceError

Output 9:
console.log(a, b, c); // 1 2 3
console.log(x, y, z); // 4 5 6

Output 10:
// 4 5 6
// 2 4 6

Output 11:
// 1 2
// 1 undefined
// undefined undefined
// 2 1
// undefined 42
// undefined undefined
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Thank you !!

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