Conditional type syntax 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 CheckType<T> = T extends number ? [1] : boolean;

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

Astring

Bnumber

Cany

Dboolean

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, otherwise boolean.

2fill in blank

medium

Complete the code to create a conditional type that returns 'true' if T is assignable to string, otherwise 'false'.

Typescript

type IsString<T> = T extends [1] ? true : false;

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

Astring

Bnumber

Cboolean

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'number' instead of 'string' in the condition.

Swapping true and false branches.

The conditional type checks if T extends string. If yes, it returns true, otherwise false.

3fill in blank

hard

Fix the error in the conditional type that should return 'never' if T is 'null', otherwise T.

Typescript

type NonNull<T> = T extends [1] ? never : T;

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Avoid

Bundefined

Cnull

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'undefined' instead of 'null' in the condition.

Returning 'never' in the wrong branch.

The conditional type checks if T extends null. If yes, it returns never, otherwise it returns T.

4fill in blank

hard

Fill both blanks to create a conditional type that returns 'Array<U>' if T is an array of U, otherwise T.

Typescript

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

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AArray<infer U>

BU

CT

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Not using 'infer' keyword.

Using 'T' instead of 'U' in the second blank.

This conditional type uses infer U to extract the element type from the array T. If T is an array, it returns Array<U>, otherwise it returns T.

5fill in blank

hard

Fill the two blanks to create a conditional type that returns the type of the property K in T if it exists, otherwise 'never'.

Typescript

type PropType<T, K extends keyof any> = K extends keyof T ? T[1] : [2];

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A[K]

Bnever

C.

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Using dot notation instead of bracket notation.

Returning 'any' instead of 'never' in the false branch.

This conditional type checks if K is a key of T. If yes, it returns the type of property K in T using T[K]. Otherwise, it returns never.