Question

Consider the following code. What will the output be, and why?

(function () {
try {
throw new Error();
} catch (x) {
var x = 1, y = 2;
(x);
}
(x);
(y);
})();

Answer 1

The JavaScript function's output will be 'undefined' for x outside of the catch block due to variable hoisting, and a 'ReferenceError' for y because y is not declared outside the catch block.

Step-by-step explanation:

The provided code is a self-invoking JavaScript function that includes a try-catch block. The try block contains a throw new Error() statement which causes an exception. The control flow then moves to the catch block where variable x is caught and subsequently re-declared with the var keyword inside the catch block, along with a declaration of another variable y. Both x and y are then assigned the values 1 and 2, respectively. Upon catching the error, the values of x and y within the catch block would be '1' and '2', respectively. However, the (x) and (y) expressions outside of the catch block will produce different results due to variable scope.

Since x is declared with the var keyword, it is hoisted to the function scope, making x accessible outside of the catch block, but with an undefined value, because var variable assignment doesn't persist across block scopes. The variable y is not accessible outside of the catch block since it's not declared outside the block, so trying to access it will result in a 'ReferenceError'.

Hence, the output of the code when the (x) and (y) outside of the catch block are executed will be 'undefined' for x, and a 'ReferenceError' will be thrown for y, causing the function to stop execution at the point where y is referenced.