Why setTimeout(fn, 0) Does NOT Run Immediately in JavaScript ?

Many JavaScript beginners believe that setTimeout(fn, 0) means the function will run immediately.

But that is not true.

Let’s understand this with a very simple example and a step-by-step explanation.

📌 The Code (Quiz Question)

console.log("A");

setTimeout(() => {
  console.log("B");
}, 0);

console.log("C");

❓ Question:

What will be the output?

A) A B C
B) A C B
C) B A C
D) A C

✅ Correct Answer: B) A C B

🧠 Step-by-Step Explanation (Very Important)

To understand this, you need to know how JavaScript executes code.

JavaScript has:

  • Call Stack
  • Task Queue (Callback Queue)
  • Event Loop

Don’t worry — we’ll keep it simple.

🔹 Step 1: Synchronous Code Runs First

JavaScript executes synchronous code line by line.

console.log("A");

➡️ "A" is printed immediately.

🔹 Step 2: setTimeout is NOT Executed Immediately

setTimeout(() => {
  console.log("B");
}, 0);

Even though the delay is 0, JavaScript does NOT execute this immediately.

Instead:

  • The callback is sent to the Task Queue
  • It waits there until the call stack is empty

❗ 0 means minimum delay, not instant execution

🔹 Step 3: Next Synchronous Line Executes

console.log("C");

➡️ "C" is printed immediately.

So far, output is:

A
C

🔹 Step 4: Event Loop Takes Control

Now:

  • Call stack is empty
  • Event loop checks the task queue
  • Executes the setTimeout callback
console.log("B");

➡️ "B" is printed last

🧾 Final Output

A
C
B

✔️ Correct option: B

⚠️ Common Mistake (Very Important)

❌ Many developers think:

setTimeout(fn, 0) = execute immediately

✅ Reality:

setTimeout(fn, 0) = execute after all synchronous code finishes

🎯 Interview Tip

If asked:

“Why does setTimeout(0) execute later?”

Say:

Because JavaScript executes synchronous code first, and setTimeout callbacks are placed in the task queue and executed only after the call stack is empty.

Leave a comment