Question 1

Take a look at the following code:

1    let x = 1;
2    function f1()
3    {
4        let x = 2;
5        console.log(x);
6    }
7    console.log(x);

Explain why line 4 and line 6 output different numbers.

My answer: Line 4 creates a new local variable x inside the function f1, which shadows the outer variable x. So, when console.log(x) executes on line 5, it refers to the local x with a value of 2. Line 7 refers to the outer variable x, which retains its original value of 1.

Question 2

Take a look at the following code:

let x = 10;

function f1() {
  console.log(x);
  let y = 20;
}

console.log(f1());
console.log(y);

What will be the output of this code. Explain your answer in 50 words or less. The code output:

10 undefined

Explain for my answer :

console.log(x) inside f1() prints the value of the outer variable x, which is 10. y is defined locally within f1(), so it's not accessible outside the function, resulting in undefined.

Question 3

Take a look at the following code:

const x = 9;

function f1(val) {
  val = val + 1;
  return val;
}

f1(x);
console.log(x);

const y = { x: 9 };

function f2(val) {
  val.x = val.x + 1;
  return val;
}

f2(y);
console.log(y);

answer Q3 THE OUTPUT :

9 { x: 10 }

MY EXPLAIN In the first case, x is a primitive (number), so passing it to f1() doesn't change its value. In the second case, y is an object, and since objects are passed by reference, modifying val.x inside f2() affects the original object y.

CodeYourFuture / Module-JS3

NW6 | Nazanin Saedi | JS3 | code-reading | Module-JS3 | week3 #318

Question 2

Explain for my answer :

Question 3

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