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;

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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];

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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;

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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;

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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;

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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.