Conditional types for overload replacement in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a conditional type that returns string if T is number, otherwise boolean.

Typescript

type Result<T> = T extends number ? [1] : boolean;

Drag options to blanks, or click blank then click option'

Aboolean

Bstring

Cnumber

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'number' instead of 'string' as the true branch.

Confusing the order of true and false branches.

The conditional type checks if T extends number; if yes, it returns string, else boolean.

2fill in blank

medium

Complete the code to create a type that returns Promise<T> if T is not already a Promise.

Typescript

type EnsurePromise<T> = T extends Promise<any> ? T : [1];

Drag options to blanks, or click blank then click option'

Avoid

BT

CPromise<T>

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Returning T directly without wrapping in Promise.

Using 'any' or 'void' incorrectly.

If T is already a Promise, return T; otherwise wrap T in Promise.

3fill in blank

hard

Fix the error in the conditional type that should extract the return type of a function type F.

Typescript

type ReturnType<F> = F extends (...args: any[]) => [1] ? [1] : never;

Drag options to blanks, or click blank then click option'

Aunknown

Bany

Cvoid

Dinfer R

Attempts:

3 left

💡 Hint

Common Mistakes

Using concrete types like 'any' or 'void' instead of 'infer R'.

Not using 'infer' keyword.

Using infer R allows extracting the return type from the function type.

4fill in blank

hard

Fill both blanks to create a type that returns the element type of an array type T, or T itself if not an array.

Typescript

type ElementType<T> = T extends [1] ? [2] : T;

Drag options to blanks, or click blank then click option'

AArray<infer U>

Binfer U

CT[]

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'T[]' instead of 'Array' in the extends clause.

Not using 'infer' to extract the element type.

The conditional checks if T is an array type and extracts its element type using infer U.

5fill in blank

hard

Fill all three blanks to define a type that converts a function type F to a version returning Promise of the original return type.

Typescript

type Asyncify<F> = F extends (...args: infer A) => [1] ? (...args: A) => [2]<[3]> : never;

Drag options to blanks, or click blank then click option'

Ainfer R

BPromise

CR

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Not using 'infer R' to extract the return type.

Using 'void' instead of 'Promise'.

Not wrapping the return type in Promise.

The type extracts the return type R, then returns a function with the same arguments but returning Promise.

Practice

1. What is the main purpose of using conditional types in TypeScript for overload replacement?

easy

A. To enforce runtime type checking in JavaScript

B. To create multiple functions with the same name but different parameters

C. To select a type based on a condition, replacing multiple overloads with one type

D. To convert all types to strings automatically

Solution

1. Step 1: Understand conditional types

   Conditional types allow choosing a type based on a condition, like an if-else for types.

2. Step 2: Relate to overload replacement

   They replace multiple function overloads by selecting the correct return type based on input types.

3. Final Answer:

   To select a type based on a condition, replacing multiple overloads with one type -> Option C

4. Quick Check:

   Conditional types = overload replacement [OK]

Hint: Think 'if-else' but for types to replace overloads [OK]

Common Mistakes:

• Confusing conditional types with runtime checks
• Thinking conditional types create multiple functions
• Believing conditional types convert types to strings

2. Which of the following is the correct syntax for a conditional type in TypeScript?

easy

A. type Result = if (T extends U) { X } else { Y };

B. type Result = T == U ? X : Y;

C. type Result = T ? U : V;

D. type Result = T extends U ? X : Y;

Solution

1. Step 1: Recall conditional type syntax

   Conditional types use the syntax: type Result = T extends U ? X : Y;

2. Step 2: Compare options

   type Result = T extends U ? X : Y; matches the correct syntax. Options A, C, and D use invalid or incorrect syntax.

3. Final Answer:

   type Result = T extends U ? X : Y; -> Option D

4. Quick Check:

   Conditional type syntax = type Result = T extends U ? X : Y; [OK]

Hint: Remember: 'extends' with '?' and ':' for conditional types [OK]

Common Mistakes:

• Using if-else syntax inside type definitions
• Confusing ternary operator with conditional types
• Using '==' instead of 'extends' keyword

3. Given the code below, what is the type of result?

type IsString<T> = T extends string ? "Yes" : "No";

let result: IsString<number>;

medium

A. "Yes"

B. "No"

C. number

D. string

Solution

1. Step 1: Analyze the conditional type

   The type IsString<T> returns "Yes" if T extends string, else "No".

2. Step 2: Apply to number

   Since number does not extend string, the type is "No".

3. Final Answer:

   "No" -> Option B

4. Quick Check:

   number extends string? No [OK]

Hint: Check if type extends string; if not, result is "No" [OK]

Common Mistakes:

• Assuming number extends string
• Confusing type with value
• Ignoring the conditional type logic

4. Identify the error in this conditional type usage:

type Check<T> = T extends number ? "Num" : "NotNum";

function test<T>(value: T): Check<T> {
  if (typeof value === "number") {
    return "Num";
  } else {
    return "NotNum";
  }
}

medium

A. Function return type does not match because conditional types are not resolved at runtime

B. The conditional type syntax is incorrect

C. The function parameter should be typed as number

D. The if condition should check for string instead of number

Solution

1. Step 1: Understand conditional types vs runtime

   Conditional types exist only at compile time and cannot be used to enforce runtime return types directly.

2. Step 2: Check function return

   The function tries to return a value matching Check<T>, but TypeScript cannot guarantee this at runtime, causing a type error.

3. Final Answer:

   Function return type does not match because conditional types are not resolved at runtime -> Option A

4. Quick Check:

   Conditional types = compile-time only [OK]

Hint: Remember conditional types are compile-time, not runtime checks [OK]

Common Mistakes:

• Thinking conditional types work at runtime
• Misusing conditional type syntax
• Changing parameter type unnecessarily

5. How can conditional types replace multiple overloads in this example?

function format(value: string): string;
function format(value: number): string;
function format(value: any): string {
  return value.toString();
}

Choose the best conditional type signature to replace these overloads.

hard

A. function format<T>(value: T): T extends string | number ? string : never;

B. function format<T>(value: T): string;

C. function format(value: T): T extends string ? string : string;

D. function format<T>(value: T): T extends string ? string : number;

Solution

1. Step 1: Understand original overloads

   Both string and number inputs return a string output.

2. Step 2: Choose conditional type matching overloads

   function format<T>(value: T): T extends string | number ? string : never; restricts T to string or number and returns string for those, else never (disallowed).

3. Step 3: Check other options

   function format<T extends string | number>(value: T): T extends string ? string : string;'s return type is redundant; function format<T>(value: T): string; allows any type; function format<T>(value: T): T extends string ? string : number; returns number incorrectly.

4. Final Answer:

   function format<T>(value: T): T extends string | number ? string : never; -> Option A

5. Quick Check:

   Conditional type restricts inputs and return string [OK]

Hint: Use conditional type to restrict inputs and unify return type [OK]

Common Mistakes:

• Allowing any input type without restriction
• Returning wrong types for some inputs
• Ignoring the need to exclude unsupported types