Event loop overview in Javascript - Step-by-Step Execution

Concept Flow - Event loop overview

Start: JS code runs

↓

Call stack executes sync code

↓

Async task encountered?

↓

Send async task to Web APIs

↓

Async task completes

↓

Callback added to task queue

↓

Event loop checks call stack

↓

If call stack empty, move callback to call stack

↓

Callback executes

↓

Repeat until all tasks done

JavaScript runs code on a call stack, sends async tasks to Web APIs, and uses the event loop to move callbacks back to the stack when ready.

Execution Sample

Javascript

console.log('Start');
setTimeout(() => console.log('Timeout'), 0);
console.log('End');

This code logs 'Start', schedules a timeout callback, then logs 'End'. The timeout callback runs after the sync code.

Execution Table

Step	Call Stack	Web APIs	Task Queue	Event Loop Action	Output
1	console.log('Start')			Execute sync code	Start
2				console.log('Start') done, removed from stack
3	setTimeout(callback, 0)	Timer started		Schedule async task
4		Timer counting down
5	console.log('End')			Execute sync code	End
6				console.log('End') done, removed from stack
7		Timer done	Callback ready	Event loop checks stack empty
8	callback()			Move callback to call stack
9	console.log('Timeout')			Execute callback	Timeout
10				callback done, removed from stack
11				All tasks done, program ends

💡 All synchronous code executed and all callbacks processed from the task queue.

Variable Tracker

Variable	Start	After Step 3	After Step 7	Final
Call Stack	Empty	setTimeout(callback, 0)	callback()	Empty
Web APIs	Empty	Timer started	Timer done	Empty
Task Queue	Empty	Empty	Callback ready	Empty
Output		Start	End	Timeout

Key Moments - 2 Insights

Why does 'Timeout' print after 'End' even though setTimeout delay is 0?

Because setTimeout callback goes to the task queue and waits until the call stack is empty (see steps 5-9 in execution_table).

What happens if the call stack is not empty when the async task finishes?

The event loop waits and does not move the callback to the call stack until it becomes empty (see step 7 and 8).

Visual Quiz - 3 Questions

Look at the execution table, what is on the call stack at step 3?

A console.log('Start')

B setTimeout(callback, 0)

C callback()

D console.log('End')

💡 Hint

Check the 'Call Stack' column at step 3 in the execution_table.

At which step does the event loop move the callback from the task queue to the call stack?

A Step 8

B Step 5

C Step 7

D Step 9

💡 Hint

Look for 'Move callback to call stack' in the 'Event Loop Action' column.

If the call stack was busy at step 7, when would the callback run?

A Before the synchronous code finishes

B Immediately at step 7

C After the call stack becomes empty

D Never

💡 Hint

Refer to the explanation in key_moments about event loop waiting for empty call stack.

Concept Snapshot

JavaScript runs synchronous code on the call stack.
Async tasks go to Web APIs and then to the task queue.
The event loop moves callbacks to the call stack when it's empty.
This ensures non-blocking, ordered execution of async code.

Full Transcript

JavaScript executes code on a single call stack. When it encounters asynchronous tasks like setTimeout, it sends them to Web APIs to wait. Once the async task finishes, its callback is placed in the task queue. The event loop constantly checks if the call stack is empty. If it is, it moves the callback from the task queue to the call stack to execute it. This process allows JavaScript to handle asynchronous operations without blocking the main thread. In the example, 'Start' and 'End' print first because they are synchronous. The 'Timeout' callback runs last, even with zero delay, because it waits for the call stack to clear.