Review the following code... console as a result of its execution?

Post author:InfraExam
Post category:Blog
Post published:October 28, 2023
Reading time:6 mins read
Post last modified:June 12, 2024

Review the following code:

let x = 100;
if (x < 100)
    x = 20;
console.log(x)

What will be displayed in the console as a result of its execution?

nothing
100
20
false

Explanation & Hint:

Here’s the breakdown of the code:

let x = 100; – Declares a variable x and initializes it with the value 100.if (x < 100) – Checks if the value of x is less than 100. However, since x is equal to 100, the condition evaluates to false.

x = 20; – Inside the if block, assigns the value 20 to x. However, since the condition in the if statement is false, this line of code will not be executed.
console.log(x); – Prints the value of x to the console. Since x is still 100, it will be displayed as 100 in the console.

The value of x remains unchanged because the if statement condition is not met, and the assignment x = 20 is not executed.

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Review the following code:

let x = 10; 
function test(x) {
    console.log(x);
}
test(20);

What will be displayed in the console as a result of its execution?

Nothing will show up.
10
20
x

Explanation & Hint:

Let’s go through the code step by step:

A variable x is declared and initialized with the value 10 using the let keyword in the global scope.
A function named test is defined with a parameter also named x. This parameter x is a local variable within the function scope, which shadows the outer variable x.
Inside the test function, the value of the local variable x (which is the parameter) is printed to the console using console.log(x).
The function definition is complete.
The test function is called with the argument 20.

When the test function is called with the argument 20, the local variable x within the function scope takes on the value of 20. Consequently, when console.log(x) is executed inside the function, it prints 20 to the console.
Therefore, the result of executing the code will be 20 displayed in the console.

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Review the following code:

let x = 10;
let y = 20;
function test(y) {
     console.log(y);
}
test(x);

What will be displayed in the console as a result of its execution?

Nothing, because the function expects the y variable to be passed and receives x instead.
20
y
10

Explanation & Hint:

The code snippet will display the value 10 in the console.

Let’s go through the code step by step:

A variable x is declared and initialized with the value 10 using the let keyword in the global scope.
A variable y is declared and initialized with the value 20 using the let keyword in the global scope.
A function named test is defined with a parameter also named y. This parameter y is a local variable within the function scope, which shadows the outer variable y.
Inside the test function, the value of the local variable y (which is the parameter) is printed to the console using console.log(y).
The function definition is complete.
The test function is called with the argument x.

When the test function is called with the argument x, the local variable y within the function scope takes on the value of the argument, which is the value of x. In this case, x is 10, so the local variable y becomes 10. Consequently, when console.log(y) is executed inside the function, it prints 10 to the console.

Therefore, the result of executing the code will be 10 displayed in the console.