counter = 0;
const logCounter = () => {
console.log(counter);
}
logCounter();
setTimeout(logCounter, 2100);
setInterval() => {
counter++;
logCounter();
},1000);

First, synchronous code is executed, and the initial logged value will be 0. Then, the event queue is visited and the first executed code will be setInterval since it has a shorter delay (1000ms), and the second logged value will be 1. After 2000ms, the second interval which will give us 2. Finally, the timeout function, which requires 2100ms to execute, will be executed and will log 2.

Order: 0 1 2 2
Under these circumstances answer is B

Correct answer D 0112