AI solution for JavaScript Quiz on BFE.dev
50. async await

The code outputs the following:

4 
1
3
6
8
2
7
5

Explanation:

• The first console.log(4) is not inside any function, so it gets executed immediately and outputs 4.
• When async1() is called, it logs 1, then it waits for async2() to finish before continuing.
• When async2() is called by async1(), it logs 3.
• While async1() is waiting for async2() to finish, the promise new Promise(...) is initiated, which logs 6 synchronously.
• After new Promise(...) is initiated, console.log(8) is executed synchronously and outputs 8.
• When async2() finishes, async1() continues with the next line and logs 2.
• Then, the then() method of the Promise is called, which logs 7.
• Finally, the setTimeout() function is called with a delay of 0 and logs 5. This may seem unexpected, but it's because the setTimeout() callback is added to the event queue and will only be executed after the synchronous tasks are finished.