The in operator narrowing in Typescript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of type narrowing with 'in' operator

What is the output of this TypeScript code when run with Node.js (ignoring type errors)?

Typescript

type Fish = { swim: () => string };
type Bird = { fly: () => string };

function move(animal: Fish | Bird) {
  if ('swim' in animal) {
    return animal.swim();
  } else {
    return animal.fly();
  }
}

const fish: Fish = { swim: () => "Swimming" };
const bird: Bird = { fly: () => "Flying" };

console.log(move(fish));
console.log(move(bird));

A Runtime error: animal.swim is not a function

B "Flying" followed by "Swimming"

C "Swimming" followed by "Flying"

D Compilation error due to type mismatch

💡 Hint

The 'in' operator checks if a property exists on the object, helping TypeScript narrow the type.

The 'in' operator narrows the union type Fish | Bird to Fish when 'swim' is present, so animal.swim() runs. Otherwise, it calls animal.fly(). The output is 'Swimming' then 'Flying'.

🧠 Conceptual

intermediate

Understanding 'in' operator narrowing with optional properties

Given the types below, which statement about the 'in' operator narrowing is true?

Typescript

type A = { x?: number };
type B = { y: string };

function test(obj: A | B) {
  if ('x' in obj) {
    // What can we say about obj here?
  }
}

A Inside the if block, obj is narrowed to type A, but 'x' may still be undefined.

B Inside the if block, obj is guaranteed to have property 'x' defined and not undefined.

C Inside the if block, obj is narrowed to type B because 'x' is optional.

D The 'in' operator cannot be used to narrow types with optional properties.

💡 Hint

The 'in' operator checks if the property exists, but optional properties may be undefined.

The 'in' operator narrows obj to type A because only A has property 'x'. However, since 'x' is optional, it may still be undefined inside the block.

🔧 Debug

advanced

Why does this 'in' operator narrowing fail?

Consider this TypeScript code snippet. Why does the compiler complain about accessing 'fly'?

Typescript

type Fish = { swim: () => void };
type Bird = { fly: () => void };

function move(animal: Fish | Bird) {
  if ('swim' in animal) {
    animal.fly();
  } else {
    animal.swim();
  }
}

A Because 'fly' is not a property of Fish, the 'in' operator narrows animal to Bird, so animal.fly() is valid.

B The compiler complains because 'fly' is not guaranteed to exist on animal even after the 'in' check.

C The 'in' operator does not narrow union types when the property is not common to all types.

D The compiler error occurs because 'swim' is missing in the else block.

💡 Hint

Check how TypeScript narrows types with the 'in' operator and property existence.

The 'in' operator narrows animal to Fish in the if block, so animal.fly() is invalid. The compiler complains because after narrowing to Fish, 'fly' does not exist.

📝 Syntax

advanced

Which option correctly uses 'in' operator for narrowing?

Which of the following TypeScript code snippets correctly narrows the type using the 'in' operator?

A if ('swim' in animal) { animal.swim(); } else { animal.fly(); }

B if (animal in 'swim') { animal.swim(); } else { animal.fly(); }

C if ('swim' of animal) { animal.swim(); } else { animal.fly(); }

D if ('swim' on animal) { animal.swim(); } else { animal.fly(); }

💡 Hint

Remember the correct syntax for the 'in' operator in TypeScript and JavaScript.

The correct syntax is 'propertyName' in object. Only option A uses this syntax correctly.

🚀 Application

expert

Determine the number of keys in the narrowed object

Given the following TypeScript code, how many keys does the object 'obj' have inside the if block?

Typescript

type A = { a: number };
type B = { b: string };

function check(obj: A | B) {
  if ('a' in obj) {
    return Object.keys(obj).length;
  }
  return 0;
}

const result = check({ a: 1, extra: true } as A & { extra: boolean });

A 0

B 1

C 3

D 2

💡 Hint

Consider how Object.keys counts enumerable own properties regardless of type narrowing.

The object passed has keys 'a' and 'extra', so Object.keys returns 2 keys inside the if block.

Practice

1.

What does the in operator do in TypeScript?

easy

A. Adds a new property to an object

B. Checks if a property exists in an object and narrows the type

C. Deletes a property from an object

D. Checks if a value is equal to a property

Solution

1. Step 1: Understand the in operator purpose

   The in operator checks if a given property exists in an object.

2. Step 2: Recognize its use in TypeScript narrowing

   TypeScript uses this check to narrow the type safely based on the presence of the property.

3. Final Answer:

   Checks if a property exists in an object and narrows the type -> Option B

4. Quick Check:

   in operator = property check and type narrowing [OK]

Hint: Remember: in checks property presence to narrow types [OK]

Common Mistakes:

• Thinking in adds or deletes properties
• Confusing in with equality checks
• Assuming in works only for arrays

2.

Which of the following is the correct syntax to check if property name exists in object obj using the in operator?

?

easy

A. "name" in obj

B. obj in "name"

C. obj.hasOwnProperty("name")

D. obj.name in

Solution

1. Step 1: Recall the in operator syntax

   The correct syntax is: "propertyName" in object.

2. Step 2: Match the syntax with options

   "name" in obj matches the correct syntax exactly: "name" in obj.

3. Final Answer:

   "name" in obj -> Option A

4. Quick Check:

   Syntax is "property" in object [OK]

Hint: Syntax is always "property" in object [OK]

Common Mistakes:

• Reversing the order of property and object
• Using method calls instead of in
• Leaving in without operands

3.

What will be the output of the following TypeScript code?

type Fish = { swim: () => void };
type Bird = { fly: () => void };

function move(animal: Fish | Bird) {
  if ("swim" in animal) {
    return "Fish can swim";
  } else {
    return "Bird can fly";
  }
}

console.log(move({ swim: () => {} }));

medium

A. Type error at compile time

B. "Bird can fly"

C. "Fish can swim"

D. Runtime error

Solution

1. Step 1: Analyze the input object

   The object passed has a swim method, so it matches the Fish type.

2. Step 2: Check the in operator condition

   The condition "swim" in animal is true, so the function returns "Fish can swim".

3. Final Answer:

   "Fish can swim" -> Option C

4. Quick Check:

   Property "swim" exists, so output is "Fish can swim" [OK]

Hint: Check which property exists to decide output [OK]

Common Mistakes:

• Confusing which property belongs to which type
• Assuming both properties exist simultaneously
• Expecting compile or runtime errors

4.

Identify the error in this TypeScript code using the in operator and fix it:

type Cat = { meow: () => void };
type Dog = { bark: () => void };

function speak(animal: Cat | Dog) {
  if (animal in "meow") {
    return "Cat speaks";
  } else {
    return "Dog speaks";
  }
}

medium

A. The in operator operands are reversed; should be "meow" in animal

B. Missing return type annotation

C. Function should use typeof instead of in

D. No error; code is correct

Solution

1. Step 1: Check in operator syntax

   The syntax requires the property name as the left operand and the object as the right operand.

2. Step 2: Identify the error in the code

   The code uses animal in "meow", which reverses operands. It should be "meow" in animal.

3. Final Answer:

   The in operator operands are reversed; should be "meow" in animal -> Option A

4. Quick Check:

   Property name must be left, object right in in [OK]

Hint: Remember: "property" in object, not object in "property" [OK]

Common Mistakes:

• Swapping operands of the in operator
• Using typeof instead of in for property checks
• Ignoring TypeScript narrowing benefits

5.

Given the following types and function, what will describeAnimal return when called with { fly: () => {}, name: "Eagle" }?

type Fish = { swim: () => void };
type Bird = { fly: () => void; name: string };

function describeAnimal(animal: Fish | Bird) {
  if ("swim" in animal) {
    return "Fish";
  } else if ("name" in animal) {
    return `Bird named ${animal.name}`;
  } else {
    return "Unknown";
  }
}

hard

A. "Fish"

B. Type error because of missing property

C. "Unknown"

D. "Bird named Eagle"

Solution

1. Step 1: Check the input object properties

   The object has fly and name properties, but no swim.

2. Step 2: Evaluate the in operator conditions

   First condition "swim" in animal is false. Second condition "name" in animal is true, so it returns Bird named Eagle.

3. Final Answer:

   "Bird named Eagle" -> Option D

4. Quick Check:

   Property "name" exists, so output is "Bird named Eagle" [OK]

Hint: Check properties in order to find matching type [OK]

Common Mistakes:

• Assuming presence of fly means Fish
• Ignoring the second in check for name
• Expecting errors due to missing swim