Type narrowing with typeof in Typescript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of type narrowing with typeof in conditional

What is the output of this TypeScript code when run with input = 42?

Typescript

function checkType(input: string | number) {
  if (typeof input === 'string') {
    return input.toUpperCase();
  } else {
    return input + 10;
  }
}
console.log(checkType(42));

A 42

B TypeError at runtime

C 52

D undefined

💡 Hint

Think about what happens when input is a number and you add 10.

The function checks if input is a string. Since 42 is a number, it adds 10 to it, resulting in 52.

❓ Predict Output

intermediate

Type narrowing effect on union types

What will this code output when value = true?

Typescript

function process(value: string | boolean) {
  if (typeof value === 'boolean') {
    return value ? 'Yes' : 'No';
  } else {
    return value.length;
  }
}
console.log(process(true));

A true

B TypeError: value.length is not a function

C "true"

D "Yes"

💡 Hint

Check the typeof condition and what happens for booleans.

Since value is boolean true, the function returns 'Yes'.

🔧 Debug

advanced

Identify the runtime error caused by incorrect type narrowing

What error does this code produce when data = 123?

Typescript

function formatData(data: string | number) {
  if (typeof data === 'string') {
    return data.trim();
  } else if (typeof data === 'number') {
    return data.toFixed(2);
  }
  return data;
}
console.log(formatData(123));

A undefined

B 123.00

C SyntaxError

D TypeError: data.trim is not a function

💡 Hint

Check the typeof checks and methods called on each type.

The code correctly narrows the type and calls toFixed on number 123, outputting '123.00'.

❓ Predict Output

advanced

Output when typeof narrowing misses a case

What is the output of this code when input = null?

Typescript

function checkInput(input: string | null) {
  if (typeof input === 'string') {
    return input.length;
  } else {
    return 'No string';
  }
}
console.log(checkInput(null));

A "No string"

B TypeError: Cannot read property 'length' of null

C 0

D null

💡 Hint

Remember typeof null is 'object', not 'string'.

Since input is null, typeof input is 'object', so the else branch runs returning 'No string'.

🧠 Conceptual

expert

Why typeof narrowing is safer than direct property access

Why is using typeof to narrow types safer than directly accessing properties without checks in TypeScript?

A Because typeof checks prevent runtime errors by ensuring the variable is the expected type before accessing properties or methods.

B Because typeof changes the variable type permanently in the program.

C Because direct property access always works regardless of type.

D Because typeof automatically converts all types to strings.

💡 Hint

Think about what happens if you try to call a string method on a number.

Using typeof ensures the code only accesses properties or methods valid for that type, avoiding runtime errors.

Practice

1. What does the typeof operator do in TypeScript?

easy

A. It converts a variable to a string.

B. It changes the type of a variable.

C. It declares a new type.

D. It checks the type of a variable at runtime.

Solution

1. Step 1: Understand the purpose of typeof

   The typeof operator is used to check the type of a variable during the program's execution (runtime).

2. Step 2: Differentiate from other options

   It does not change, declare, or convert types; it only inspects the type.

3. Final Answer:

   It checks the type of a variable at runtime. -> Option D

4. Quick Check:

   typeof checks type at runtime [OK]

Hint: Remember: typeof tells you the type, it doesn't change it. [OK]

Common Mistakes:

• Thinking typeof changes the variable's type
• Confusing typeof with type declaration
• Assuming typeof converts values to strings

2. Which of the following is the correct syntax to narrow a variable value to a string using typeof?

easy

A. if (typeof value === 'string') { }

B. if (typeof value = 'string') { }

C. if (typeof value == string) { }

D. if (typeof value === string) { }

Solution

1. Step 1: Check correct equality operator and string literal

   The correct way to compare is using triple equals === and the type name as a string literal, like 'string'.

2. Step 2: Identify syntax errors in other options

   if (typeof value = 'string') { } uses single equals which is assignment, not comparison. Options A and C miss quotes around 'string'.

3. Final Answer:

   if (typeof value === 'string') { } -> Option A

4. Quick Check:

   Use === and quotes for typeof checks [OK]

Hint: Use === and quotes around type names in typeof checks. [OK]

Common Mistakes:

• Using single = instead of ===
• Forgetting quotes around type names
• Using == instead of ===

3. What will be the output of this TypeScript code?

function checkType(x: string | number) {
  if (typeof x === 'string') {
    return x.toUpperCase();
  } else {
    return x + 10;
  }
}

console.log(checkType('hello'));
console.log(checkType(5));

medium

A. HELLO and 510

B. hello and 15

C. HELLO and 15

D. Error at runtime

Solution

1. Step 1: Analyze the input 'hello'

   Since 'hello' is a string, typeof x === 'string' is true, so x.toUpperCase() returns 'HELLO'.

2. Step 2: Analyze the input 5

   Since 5 is a number, the else block runs, returning 5 + 10 = 15.

3. Final Answer:

   HELLO and 15 -> Option C

4. Quick Check:

   String upper + number addition = HELLO, 15 [OK]

Hint: Check typeof to pick string or number logic correctly. [OK]

Common Mistakes:

• Concatenating number and string instead of adding
• Calling string methods on numbers
• Confusing toUpperCase output

4. Identify the error in this TypeScript code using typeof for type narrowing:

function process(value: string | boolean) {
  if (typeof value === 'boolean') {
    return value.toUpperCase();
  } else {
    return value ? 'Yes' : 'No';
  }
}

medium

A. Calling toUpperCase on a boolean value

B. Using typeof with boolean type

C. Missing return type annotation

D. Incorrect if-else structure

Solution

1. Step 1: Check the type narrowing condition

   The condition typeof value === 'boolean' correctly narrows value to boolean.

2. Step 2: Identify method call on wrong type

   Inside this block, calling value.toUpperCase() is invalid because booleans do not have this method.

3. Final Answer:

   Calling toUpperCase on a boolean value -> Option A

4. Quick Check:

   Boolean has no toUpperCase method [OK]

Hint: Check method calls match the narrowed type. [OK]

Common Mistakes:

• Assuming booleans have string methods
• Ignoring type narrowing effect
• Confusing boolean and string logic

5. You have a function that accepts input: string | number | boolean. You want to return:
- The length of the string if input is a string,
- The number multiplied by 2 if input is a number,
- The string 'True' or 'False' if input is a boolean.

Which of the following implementations correctly uses typeof for type narrowing?

hard

A. function handle(input: string | number | boolean) {
if (typeof input === 'boolean') {
return input.length;
} else if (typeof input === 'string') {
return input * 2;
} else {
return input ? 'True' : 'False';
}
}

B. function handle(input: string | number | boolean) {
if (typeof input === 'string') {
return input.length;
} else if (typeof input === 'number') {
return input * 2;
} else {
return input ? 'True' : 'False';
}
}

C. function handle(input: string | number | boolean) {
if (typeof input === 'number') {
return input.length;
} else if (typeof input === 'boolean') {
return input * 2;
} else {
return input ? 'True' : 'False';
}
}

D. function handle(input: string | number | boolean) {
if (typeof input === 'string') {
return input.toUpperCase();
} else if (typeof input === 'number') {
return input + 2;
} else {
return input ? 'True' : 'False';
}
}

Solution

1. Step 1: Check string handling

   function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.length; } else if (typeof input === 'number') { return input * 2; } else { return input ? 'True' : 'False'; } } returns input.length for strings, which is correct. function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.toUpperCase(); } else if (typeof input === 'number') { return input + 2; } else { return input ? 'True' : 'False'; } } returns input.toUpperCase(), which is not requested.

2. Step 2: Check number handling

   function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.length; } else if (typeof input === 'number') { return input * 2; } else { return input ? 'True' : 'False'; } } multiplies number by 2, as required. function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.toUpperCase(); } else if (typeof input === 'number') { return input + 2; } else { return input ? 'True' : 'False'; } } adds 2, which is incorrect. Options B and C misuse length or multiplication on wrong types.

3. Step 3: Check boolean handling

   function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.length; } else if (typeof input === 'number') { return input * 2; } else { return input ? 'True' : 'False'; } } returns 'True' or 'False' strings correctly. Others mix up boolean logic or return wrong types.

4. Final Answer:

   function handle(input: string | number | boolean) { if (typeof input === 'string') { return input.length; } else if (typeof input === 'number') { return input * 2; } else { return input ? 'True' : 'False'; } } -> Option B

5. Quick Check:

   Match typeof checks with correct operations [OK]

Hint: Match each typeof check with correct operation for that type. [OK]

Common Mistakes:

• Calling string methods on numbers or booleans
• Using length on numbers or booleans
• Mixing up operations for each type