Mapped type with template literals in Typescript - Step-by-Step Execution

Concept Flow - Mapped type with template literals

Start with base type

↓

Iterate over keys with 'keyof'

↓

Apply template literal to each key

↓

Create new mapped type with modified keys

↓

Result: type with new keys and original values

This flow shows how TypeScript creates a new type by looping over keys of an existing type and changing each key using template literals.

Execution Sample

Typescript

type Person = {
  name: string;
  age: number;
};

type PrefixedPerson = {
  [K in keyof Person as `get${Capitalize<string & K>}`]: () => Person[K];
};

This code creates a new type with keys prefixed by 'get' and capitalized, each returning the original property type as a function.

Execution Table

Step	Key (K)	Template Literal Key	Mapped Key	Value Type
1	name	`get${Capitalize<'name'>}`	getName	() => string
2	age	`get${Capitalize<'age'>}`	getAge	() => number
3	-	-	-	Mapped type complete with keys getName and getAge

💡 All keys of Person processed, mapped type created with new keys and function types.

Variable Tracker

Variable	Start	After 1	After 2	Final
K	-	name	age	-
Mapped Key	-	getName	getAge	-
Value Type	-	() => string	() => number	-

Key Moments - 2 Insights

Why do we use 'as' in the mapped type?

The 'as' keyword renames each key using the template literal. Without it, keys stay the same. See execution_table rows 1 and 2 where keys change from 'name' to 'getName'.

What does 'Capitalize<string & K>' do?

It capitalizes the first letter of each key string. For example, 'name' becomes 'Name'. This is shown in execution_table where 'name' becomes 'getName'.

Visual Quiz - 3 Questions

Look at the execution table, what is the mapped key for the original key 'age'?

A getage

B Age

C getAge

D age

💡 Hint

Check the 'Mapped Key' column in row 2 of the execution_table.

At which step does the mapped type finish creating all keys?

A Step 3

B Step 2

C Step 1

D Step 4

💡 Hint

Look at the exit note and the last row in execution_table.

If we remove 'Capitalize' from the template literal, what would the mapped key for 'name' be?

A getName

B getname

C Name

D name

💡 Hint

Refer to how 'Capitalize' changes 'name' to 'Name' in the execution_table.

Concept Snapshot

Mapped types loop over keys of a type.
Use 'as' with template literals to rename keys.
Template literals can modify keys dynamically.
Capitalize utility changes first letter uppercase.
Result is a new type with transformed keys and original values.

Full Transcript

This visual execution shows how TypeScript creates a mapped type using template literals. Starting from a base type with keys 'name' and 'age', it loops over each key. For each key, it applies a template literal that prefixes 'get' and capitalizes the key. This renames 'name' to 'getName' and 'age' to 'getAge'. The mapped type then assigns each new key a function type returning the original property type. The execution table traces each step, showing key transformation and value types. Key moments clarify why 'as' is needed to rename keys and how 'Capitalize' works. The quiz tests understanding of key renaming and the mapping process. This helps beginners see how mapped types with template literals create new types with dynamic keys.

Practice

1.

What does a mapped type with template literals in TypeScript allow you to do?

Example: { [K in keyof T as `get${Capitalize}`]: () => T[K] }

easy

A. Create new variables at runtime

B. Rename keys dynamically in object types

C. Change values of an object at runtime

D. Execute functions based on key names

Solution

1. Step 1: Understand mapped types

   Mapped types create new object types by transforming keys or values of an existing type.

2. Step 2: Role of template literals

   Template literals let you rename keys dynamically by combining strings and key names.

3. Final Answer:

   Rename keys dynamically in object types -> Option B

4. Quick Check:

   Mapped type + template literals = dynamic key renaming [OK]

Hint: Mapped types + template literals = dynamic key renaming [OK]

Common Mistakes:

• Confusing runtime behavior with type transformations
• Thinking it changes object values instead of keys
• Assuming it creates variables or runs code

2.

Which of the following is the correct syntax to rename keys using a mapped type with template literals?

type RenameKeys<T> = { [K in keyof T as ???]: T[K] }

easy

A. `new_${string & K}`

B. `new_${K}`

C. `new_${number & K}`

D. `new_${any & K}`

Solution

1. Step 1: Understand key remapping syntax

   Use as with template literals and intersect key type with string to ensure keys are strings.

2. Step 2: Check each option

   `new_${string & K}` uses string & K which is correct; others use incorrect types like number or any.

3. Final Answer:

   `new_${string & K}` -> Option A

4. Quick Check:

   Use string & K for safe key renaming [OK]

Hint: Use string & K to rename keys safely in template literals [OK]

Common Mistakes:

• Using K directly without string intersection
• Using number or any instead of string & K
• Missing backticks for template literals

3.

What is the resulting type of the following code?

type User = { name: string; age: number };
type PrefixedUser = { [K in keyof User as `get${Capitalize}`]: () => User[K] };

medium

A. { name: () => string; age: () => number }

B. { getName: string; getAge: number }

C. { getname: () => string; getage: () => number }

D. { getName: () => string; getAge: () => number }

Solution

1. Step 1: Analyze key transformation

   Keys 'name' and 'age' are capitalized and prefixed with 'get', becoming 'getName' and 'getAge'.

2. Step 2: Analyze value transformation

   Values become functions returning the original property types: () => string and () => number.

3. Final Answer:

   { getName: () => string; getAge: () => number } -> Option D

4. Quick Check:

   Mapped keys capitalized + function values = { getName: () => string; getAge: () => number } [OK]

Hint: Capitalize keys and prefix with 'get' for function types [OK]

Common Mistakes:

• Not capitalizing keys
• Returning values instead of functions
• Using original keys without prefix

4.

Identify the error in this mapped type using template literals:

type Flags = { isActive: boolean; isAdmin: boolean };
type PrefixedFlags = { [K in keyof Flags as `flag_${K}`]: Flags[K] };

medium

A. Key remapping must use Capitalize utility

B. Cannot use template literals in mapped types

C. Missing string intersection with K in template literal

D. Flags[K] is invalid syntax

Solution

1. Step 1: Check key remapping syntax

   Template literal keys must intersect K with string: string & K to ensure keys are strings.

2. Step 2: Identify missing intersection

   The code uses `flag_${K}` without string & K, causing a type error.

3. Final Answer:

   Missing string intersection with K in template literal -> Option C

4. Quick Check:

   Use string & K in template literals to fix error [OK]

Hint: Always intersect K with string in template literals [OK]

Common Mistakes:

• Omitting string & K intersection
• Thinking Capitalize is mandatory
• Believing template literals can't be used here

5.

Given this type, how would you create a mapped type that prefixes keys with on and capitalizes them, but only for keys starting with handle?

type Handlers = { handleClick: () => void; handleHover: () => void; reset: () => void };

hard

A. type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] };

B. type Prefixed = { [K in keyof Handlers as `on${Capitalize}`]: Handlers[K] };

C. type Prefixed = { [K in keyof Handlers as K extends string ? `on${Capitalize<K>}` : never]: Handlers[K] };

D. type Prefixed = { [K in keyof Handlers as K extends `on${string}` ? `handle${Capitalize}` : never]: Handlers[K] };

Solution

1. Step 1: Filter keys starting with 'handle'

   Use conditional type K extends `handle${string}` to select only keys starting with 'handle'.

2. Step 2: Rename keys with 'on' prefix and capitalize

   Use template literal `on${Capitalize}` to rename keys accordingly.

3. Step 3: Assign original value types

   Keep the original function types from Handlers[K].

4. Final Answer:

   type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; -> Option A

5. Quick Check:

   Conditional key remapping with prefix and capitalize = type Prefixed = { [K in keyof Handlers as K extends `handle${string}` ? `on${Capitalize}` : never]: Handlers[K] }; [OK]

Hint: Use conditional types to filter keys before renaming [OK]

Common Mistakes:

• Not filtering keys by prefix
• Applying rename to all keys
• Incorrect conditional syntax