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What is the output of the following code snippet?
var x = 5;
function foo() {
  console.log(x);
  var x = 10;
  function bar() {
    console.log(x);
  }
  bar();
}
foo();
  • a)
    0, 0
  • b)
    undefined, undefined
  • c)
    10, 10
  • d)
    undefined, 10
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
What is the output of the following code snippet?var x = 5;function fo...
The variable x is declared in the function foo but is not assigned a value before the first console.log statement. Therefore, it outputs undefined. Inside the bar function, it logs the value of x, which is 10.
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Community Answer
What is the output of the following code snippet?var x = 5;function fo...
Explanation:

The output of the code snippet is "undefined, 10".

- When the function `foo()` is called, the first `console.log(x)` statement is executed. However, at this point, the variable `x` is declared but not yet assigned a value. Therefore, the value of `x` is `undefined` inside the `foo()` function. So, the first `console.log(x)` statement prints `undefined`.

- Inside the `foo()` function, a new variable `x` is declared and assigned a value of `10`. This creates a new local variable `x` that shadows the global variable `x`. The global variable `x` is still undefined.

- The function `bar()` is defined inside the `foo()` function and called immediately. Inside the `bar()` function, the `console.log(x)` statement is executed. Since `x` is not declared in the `bar()` function, JavaScript looks for the variable in the scope chain. It first checks the local scope of `bar()`, where `x` is not defined. Then, it checks the scope of the parent function `foo()`, where `x` is defined. So, the `console.log(x)` statement prints `10`.

- Finally, the `foo()` function is called, and it executes the code inside it. This results in the output "undefined, 10" as explained above.

In summary, the variable `x` inside the `foo()` function is a local variable that shadows the global variable `x`. The first `console.log(x)` statement prints `undefined` because the local variable `x` is declared but not yet assigned a value. The second `console.log(x)` statement prints `10` because it refers to the local variable `x` within the `foo()` function.
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What is the output of the following code snippet?var x = 5;function foo() { console.log(x); var x = 10; function bar() { console.log(x); } bar();}foo();a)0, 0b)undefined, undefinedc)10, 10d)undefined, 10Correct answer is option 'D'. Can you explain this answer?
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