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What will be the output of the following code snippet?
let x = 1;
function foo() {
console.log(x);
let x = 2;
}
foo();
let x = 1;
function foo() {
console.log(x);
let x = 2;
}
foo();
- a)1
- b)2
- c)undefined
- d)ReferenceError
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
What will be the output of the following code snippet?let x = 1;functi...
Explanation:
The code snippet provided defines a variable `x` with the value of 1 and a function `foo`. When `foo` is called, it attempts to log the value of `x` to the console and then defines a new variable `x` with the value of 2.
Variable Hoisting:
In JavaScript, variable declarations using the `let` keyword are hoisted to the top of their scope, but they are not initialized. This means that the variable `x` is hoisted to the top of the `foo` function, but its value is not set until the line where it is defined.
Execution:
When the code is executed, the following steps occur:
1. The variable `x` is declared and initialized with the value of 1.
2. The `foo` function is defined.
3. The `foo` function is called.
4. Inside the `foo` function, the `console.log(x)` statement is executed.
- At this point, the variable `x` is hoisted to the top of the `foo` function, but its value is not set yet.
- Since the value of `x` is not set, it is undefined.
- Therefore, trying to log the value of `x` to the console will result in the output of `undefined`.
5. The variable `x` is defined with the value of 2.
- This line is never reached because the `console.log(x)` statement causes an error.
Error:
The error thrown is a ReferenceError because the variable `x` is referenced before it has been initialized with a value. This error occurs because the `let` keyword does not allow hoisting to the top of the scope, unlike the `var` keyword.
Therefore, the correct answer is option 'D) ReferenceError'.
The code snippet provided defines a variable `x` with the value of 1 and a function `foo`. When `foo` is called, it attempts to log the value of `x` to the console and then defines a new variable `x` with the value of 2.
Variable Hoisting:
In JavaScript, variable declarations using the `let` keyword are hoisted to the top of their scope, but they are not initialized. This means that the variable `x` is hoisted to the top of the `foo` function, but its value is not set until the line where it is defined.
Execution:
When the code is executed, the following steps occur:
1. The variable `x` is declared and initialized with the value of 1.
2. The `foo` function is defined.
3. The `foo` function is called.
4. Inside the `foo` function, the `console.log(x)` statement is executed.
- At this point, the variable `x` is hoisted to the top of the `foo` function, but its value is not set yet.
- Since the value of `x` is not set, it is undefined.
- Therefore, trying to log the value of `x` to the console will result in the output of `undefined`.
5. The variable `x` is defined with the value of 2.
- This line is never reached because the `console.log(x)` statement causes an error.
Error:
The error thrown is a ReferenceError because the variable `x` is referenced before it has been initialized with a value. This error occurs because the `let` keyword does not allow hoisting to the top of the scope, unlike the `var` keyword.
Therefore, the correct answer is option 'D) ReferenceError'.
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Community Answer
What will be the output of the following code snippet?let x = 1;functi...
The variable 'x' is declared using 'let' inside the function 'foo'. The 'console.log(x)' line tries to access 'x' before it is declared, resulting in a ReferenceError.
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What will be the output of the following code snippet?let x = 1;function foo() { console.log(x); let x = 2;}foo();a)1b)2c)undefinedd)ReferenceErrorCorrect answer is option 'D'. Can you explain this answer?
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