Closures in Javascript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of a closure with loop variable

What is the output of the following code snippet?

Javascript

function createFunctions() {
  const funcs = [];
  for (let i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}
const functions = createFunctions();
console.log(functions[0]());
console.log(functions[1]());
console.log(functions[2]());

A 0 0 0

B 3 3 3

C undefined undefined undefined

D 0 1 2

💡 Hint

Think about how the let keyword affects the loop variable scope inside closures.

Using let in the loop creates a new binding for each iteration, so each function remembers its own i value.

❓ Predict Output

intermediate

Closure capturing variable after loop with var

What will be printed by this code?

Javascript

function createFunctions() {
  const funcs = [];
  for (var i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}
const functions = createFunctions();
console.log(functions[0]());
console.log(functions[1]());
console.log(functions[2]());

A 3 3 3

B undefined undefined undefined

C 0 0 0

D 0 1 2

💡 Hint

Consider how var differs from let in loops and closures.

Using var declares i in function scope, so all functions share the same i which ends at 3 after the loop.

🧠 Conceptual

advanced

Why closures keep variables alive

Why do closures in JavaScript keep variables alive even after the outer function has finished?

A Because closures create a copy of the variables and store them separately.

B Because closures keep a reference to the variables in the outer function's scope, preventing garbage collection.

C Because JavaScript automatically duplicates all variables used inside functions.

D Because closures run the outer function again every time they are called.

💡 Hint

Think about how memory and references work in JavaScript functions.

Closures keep references to variables in their outer scope, so those variables stay in memory as long as the closure exists.

❓ Predict Output

advanced

Output of nested closures with parameters

What is the output of this code?

Javascript

function outer(x) {
  return function inner(y) {
    return function innermost(z) {
      return x + y + z;
    };
  };
}
const result = outer(1)(2)(3);
console.log(result);

A undefined

B 123

C 6

D NaN

💡 Hint

Add the numbers passed through each function call.

Each function returns another function that remembers the parameters passed, so the final sum is 1 + 2 + 3 = 6.

❓ Predict Output

expert

Closure with asynchronous behavior and loop

What will this code print to the console?

Javascript

for (var i = 0; i < 3; i++) {
  setTimeout(function() {
    console.log(i);
  }, 100);
}

A 3 3 3

B 0 1 2

C undefined undefined undefined

D 0 0 0

💡 Hint

Consider how var and asynchronous callbacks interact.

The loop finishes before the timeouts run, so i is 3 in all callbacks, printing 3 three times.

Practice

1. What is a closure in JavaScript?

easy

A. A function that has no parameters

B. A function that runs only once

C. A function that returns another function without variables

D. A function that remembers variables from its creation context

Solution

1. Step 1: Understand closure definition

   A closure is a function that keeps access to variables from the place where it was created, even after that place has finished running.

2. Step 2: Match definition to options

   A function that remembers variables from its creation context correctly describes this behavior, while others describe unrelated concepts.

3. Final Answer:

   A function that remembers variables from its creation context -> Option D

4. Quick Check:

   Closure = function remembering variables [OK]

Hint: Closures remember variables from where they were made [OK]

Common Mistakes:

• Thinking closures are functions without parameters
• Confusing closures with immediately invoked functions
• Believing closures only run once

2. Which of the following is the correct syntax to create a closure in JavaScript?

easy

A. function outer() { let x = 5; function inner() { return x; } }

B. function outer() { let x = 5; return x; }

C. function outer() { let x = 5; return function inner() { return x; } }

D. function outer() { return function() { let x = 5; } }

Solution

1. Step 1: Identify closure syntax

   A closure requires an inner function that accesses variables from the outer function and is returned so it can be used later.

2. Step 2: Check options for correct closure

   function outer() { let x = 5; return function inner() { return x; } } returns an inner function that uses outer variable x, forming a closure. Others either don't return the function or don't access outer variables correctly.

3. Final Answer:

   function outer() { let x = 5; return function inner() { return x; } } -> Option C

4. Quick Check:

   Return inner function accessing outer variable [OK]

Hint: Closure needs returned inner function accessing outer variable [OK]

Common Mistakes:

• Not returning the inner function
• Returning only the variable, not a function
• Defining inner function but not using outer variables

3. What will be the output of the following code?

function makeCounter() {
  let count = 0;
  return function() {
    count += 1;
    return count;
  };
}
const counter = makeCounter();
console.log(counter());
console.log(counter());

medium

A. 1 and 2

B. 0 and 1

C. undefined and undefined

D. 1 and 1

Solution

1. Step 1: Understand closure behavior in makeCounter

   The inner function increments count and returns it. The variable count is remembered by the closure.

2. Step 2: Trace the calls to counter()

   First call: count = 0 + 1 = 1, returns 1. Second call: count = 1 + 1 = 2, returns 2.

3. Final Answer:

   1 and 2 -> Option A

4. Quick Check:

   Closure keeps count incrementing: 1, 2 [OK]

Hint: Closure keeps variable state between calls [OK]

Common Mistakes:

• Thinking count resets each call
• Expecting output 1 and 1
• Confusing closure with global variable

4. Find the error in this closure example:

function createFuncs() {
  let funcs = [];
  for (var i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}
const functions = createFuncs();
console.log(functions[0]());
console.log(functions[1]());

medium

A. All functions return 3 instead of 0, 1, 2

B. Syntax error due to missing semicolons

C. Functions return undefined

D. Functions throw a ReferenceError

Solution

1. Step 1: Identify loop variable scope

   The loop uses var i, which is function-scoped, so all inner functions share the same i.

2. Step 2: Understand closure captures final i value

   After loop ends, i is 3. All functions return 3, not 0, 1, 2 as expected.

3. Final Answer:

   All functions return 3 instead of 0, 1, 2 -> Option A

4. Quick Check:

   var in loop causes shared i = 3 [OK]

Hint: var in loop shares variable; use let for block scope [OK]

Common Mistakes:

• Assuming each function captures separate i
• Thinking semicolons cause error
• Expecting ReferenceError or undefined

5. How can you fix the following code so each function remembers its own value of i?

function createFuncs() {
  let funcs = [];
  for (var i = 0; i < 3; i++) {
    funcs.push(function() { return i; });
  }
  return funcs;
}

hard

A. Use var but call the function immediately

B. Change var to let in the for loop declaration

C. Return i directly instead of a function

D. Declare i as a global variable

Solution

1. Step 1: Understand block scope with let

   Using let in the for loop creates a new binding of i for each iteration, so closures capture different i values.

2. Step 2: Compare with var behavior

   var is function-scoped, so all closures share the same i. Changing to let fixes this by giving each iteration its own i.

3. Final Answer:

   Change var to let in the for loop declaration -> Option B

4. Quick Check:

   let in loop creates separate i per iteration [OK]

Hint: Use let in loops to create separate closure variables [OK]

Common Mistakes:

• Trying to fix by making i global
• Returning i instead of function
• Using var with immediate call but misunderstanding closures