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What will be the output of the following code snippet?
let x = 5;
(function() {
console.log(x);
let x = 10;
})();
let x = 5;
(function() {
console.log(x);
let x = 10;
})();
- a)5
- b)10
- 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 = 5;(funct...
The variable 'x' is declared using 'let' inside the immediately-invoked function expression (IIFE). Since 'let' has block scope, the declaration of 'x' inside the function creates a new variable that is not accessible before the declaration.
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Community Answer
What will be the output of the following code snippet?let x = 5;(funct...
-
Explanation:
- Variable Hoisting:
- In JavaScript, variable declarations are hoisted to the top of their containing scope. This means that regardless of where a variable is declared within a function, it is moved to the top of the function during the compilation phase.
- Function Scope:
- In the given code snippet, the variable `x` is declared both globally and within the immediately invoked function expression (IIFE).
- Accessing Undefined Variable:
- When the IIFE is executed, it tries to log the value of `x` before it is actually declared within the function. This results in a ReferenceError because the variable `x` is accessed before it is initialized.
- Variable Shadowing:
- The `x` variable declared within the IIFE shadows the global `x` variable within the function scope, which is why the ReferenceError occurs when trying to access the local `x` before it is declared.
- Correct Output:
- The correct output of the code snippet is a ReferenceError, indicating that the variable `x` is accessed before it is declared within the function scope.
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What will be the output of the following code snippet?let x = 5;(function() { console.log(x); let x = 10;})();a)5b)10c)undefinedd)ReferenceErrorCorrect answer is option 'D'. Can you explain this answer?
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