Salesforce JavaScript Developer Practice Exam 2025 – 400 Free Practice Questions to Pass the Exam

The code provided defines three functions: `foo`, `bar`, and `baz`. The execution flow of these functions reveals the output order.

When `bar()` is called, it invokes a `setTimeout` function that schedules the output of 'Second' to occur after the timer (default is 0 milliseconds) has elapsed. This means that while 'Second' is scheduled, control is immediately returned to the next line of code since it's asynchronous.

Next, `foo()` is executed, which immediately outputs 'First' to the console. After that, `baz()` is called, which outputs 'Third' to the console as well.

At the end of this sequence, the scheduled function from `bar` will execute and output 'Second' to the console, but only after the synchronous operations (the calls to `foo` and `baz`) have completed.

So, in the order of execution:

1. `bar()` schedules 'Second' (but it doesn't log it immediately).

2. `foo()` logs 'First'.

3. `baz()` logs 'Third'.

4. After all synchronous code has executed, 'Second' is logged by the scheduled callback from `bar`.

Hence, the final order of outputs is 'First