Non-distributive conditional types in Typescript - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a conditional type that checks if T is a string.

Typescript

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

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' in the extends clause.

Confusing the syntax of conditional types.

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

2fill in blank

medium

Complete the code to prevent distributive behavior by wrapping T in a tuple.

Typescript

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

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

Anumber

Bany

Cstring

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Not wrapping T in a tuple, causing distributive behavior.

Using the wrong type inside the extends clause.

Wrapping T in a tuple [T] stops TypeScript from distributing the conditional type over unions.

3fill in blank

hard

Fix the error in the conditional type to avoid distribution over union types.

Typescript

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

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

Aboolean

Bstring

Cnumber

D[string]

Attempts:

3 left

💡 Hint

Common Mistakes

Using string directly, which causes distribution.

Not understanding how wrapping in tuples affects conditional types.

Using [string] inside the extends clause wraps the type in a tuple, preventing distribution.

4fill in blank

hard

Fill both blanks to create a non-distributive conditional type that checks if T is assignable to U.

Typescript

type IsAssignable<T, U> = [T] extends [[1]] ? [2] : false;

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

AU

Btrue

CT

Dfalse

Attempts:

3 left

💡 Hint

Common Mistakes

Not wrapping T in a tuple.

Returning T instead of true in the true branch.

This type checks if T is assignable to U without distribution by wrapping T in a tuple. If yes, it returns true, else false.

5fill in blank

hard

Fill all three blanks to create a non-distributive conditional type that returns the type if it extends U, else never.

Typescript

type FilterType<T, U> = [T] extends [[1]] ? [2] : [3];

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

AU

BT

Cnever

Dunknown

Attempts:

3 left

💡 Hint

Common Mistakes

Returning unknown instead of never in the false branch.

Not wrapping T in a tuple.

This type filters T by checking if it extends U without distribution. If yes, it returns T, else never.

Practice

1. What does a non-distributive conditional type in TypeScript do differently compared to a distributive conditional type?

easy

A. It converts all types to string before evaluation.

B. It always distributes the condition over each member of the union type.

C. It only works with primitive types, not unions.

D. It evaluates the condition on the entire union type as a whole, not on each member separately.

Solution

1. Step 1: Understand distributive conditional types

   Distributive conditional types apply the condition to each member of a union type separately.

2. Step 2: Understand non-distributive conditional types

   Non-distributive conditional types evaluate the condition on the entire union type as a single entity, not distributing over members.

3. Final Answer:

   It evaluates the condition on the entire union type as a whole, not on each member separately. -> Option D

4. Quick Check:

   Non-distributive = whole union check [OK]

Hint: Non-distributive means no splitting union, check whole type [OK]

Common Mistakes:

• Thinking conditional always distributes over unions
• Confusing non-distributive with type narrowing
• Assuming it only works on primitives

2. Which syntax correctly defines a non-distributive conditional type in TypeScript?

easy

A. type NonDist<T> = T | string ? true : false;

B. type NonDist<T> = T extends string ? true : false;

C. type NonDist<T> = [T] extends [string] ? true : false;

D. type NonDist<T> = T & string ? true : false;

Solution

1. Step 1: Recognize non-distributive syntax

   Wrapping the type parameter in a tuple, like [T], prevents distribution over unions.

2. Step 2: Compare options

   type NonDist<T> = [T] extends [string] ? true : false; uses [T] extends [string], which is the correct non-distributive syntax. type NonDist<T> = T extends string ? true : false; distributes over unions. Options C and D are invalid conditional syntax.

3. Final Answer:

   type NonDist<T> = [T] extends [string] ? true : false; -> Option C

4. Quick Check:

   Tuple wrapping disables distribution [OK]

Hint: Wrap type in tuple [T] to prevent distribution [OK]

Common Mistakes:

• Not wrapping type in tuple, causing distribution
• Using invalid conditional syntax
• Confusing union with intersection in condition

3. Given the code:

type Test<T> = T extends string ? 'yes' : 'no';
type Result = Test<'a' | number>;

What is the type of Result?

medium

A. 'yes' | 'no'

B. 'yes'

C. 'no'

D. Error

Solution

1. Step 1: Understand distributive conditional types

   Since T is a union 'a' | number, the condition distributes over each member.

2. Step 2: Evaluate each member

   'a' extends string is true, so 'yes'; number extends string is false, so 'no'. Combined result is 'yes' | 'no'.

3. Final Answer:

   'yes' | 'no' -> Option A

4. Quick Check:

   Distributes over union = 'yes' | 'no' [OK]

Hint: Conditional distributes over union unless wrapped [OK]

Common Mistakes:

• Assuming whole union checked at once
• Expecting only 'yes' or only 'no'
• Confusing union with intersection

4. Identify the error in this non-distributive conditional type:

type NonDist<T> = T extends [string] ? true : false;

medium

A. It incorrectly checks if T extends a tuple containing string, not preventing distribution.

B. It correctly prevents distribution by wrapping T in a tuple.

C. It causes a syntax error because of missing brackets.

D. It always returns false regardless of T.

Solution

1. Step 1: Analyze the condition

   The condition checks if T extends [string], a tuple type, not wrapping T.

2. Step 2: Understand distribution prevention

   To prevent distribution, T itself must be wrapped in a tuple: [T] extends [string]. This code does not do that, so distribution is not prevented.

3. Final Answer:

   It incorrectly checks if T extends a tuple containing string, not preventing distribution. -> Option A

4. Quick Check:

   Wrap T, not string, to prevent distribution [OK]

Hint: Wrap T in tuple, not the type to check [OK]

Common Mistakes:

• Wrapping the wrong side of extends
• Confusing tuple with type wrapper
• Assuming syntax error instead of logic error

5. How can you write a non-distributive conditional type that checks if a union type T is exactly string | number (no more, no less)?

hard

A. type Exact<T> = T extends string | number ? true : false;

B. type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false;

C. type Exact<T> = T extends [string | number] ? true : false;

D. type Exact<T> = [T] extends [string] | [number] ? true : false;

Solution

1. Step 1: Prevent distribution with tuple wrapping

   Wrapping T in a tuple [T] prevents distribution over unions.

2. Step 2: Use mutual assignability checks

   Check [T] extends [string | number] and [string | number] extends [T] to ensure exact match, no extra or missing types.

3. Final Answer:

   type Exact<T> = [T] extends [string | number] ? ([string | number] extends [T] ? true : false) : false; -> Option B

4. Quick Check:

   Tuple wrap + mutual extends = exact type check [OK]

Hint: Wrap T and check both extends directions for exact match [OK]

Common Mistakes:

• Not preventing distribution, causing partial matches
• Checking only one extends direction
• Using incorrect tuple wrapping