How TypeScript infers generic types - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to define a generic function that returns the input value.

Typescript

function identity<T>([1]: T): T {
  return [1];
}

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

Avalue

Btype

Cinput

Darg

Attempts:

3 left

💡 Hint

Common Mistakes

Using a parameter name that doesn't match the return statement.

Using a non-generic type.

The generic function uses a parameter named 'value' of type T and returns it.

2fill in blank

medium

Complete the code to call the generic function with a string argument and let TypeScript infer the type.

Typescript

const result = identity([1]);

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

A42

Bnull

Ctrue

D'hello'

Attempts:

3 left

💡 Hint

Common Mistakes

Passing a number or boolean instead of a string.

Passing null which is not inferred as a string.

Passing a string literal 'hello' lets TypeScript infer T as string.

3fill in blank

hard

Fix the error in the generic function call by specifying the type explicitly.

Typescript

const result = identity<[1]>(42);

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

Aboolean

Bstring

Cnumber

Dany

Attempts:

3 left

💡 Hint

Common Mistakes

Specifying a type that doesn't match the argument type.

Omitting the generic type when explicit specification is needed.

Explicitly specifying 'number' as the generic type matches the argument 42.

4fill in blank

hard

Fill both blanks to define a generic function that returns the length of an array of generic type T.

Typescript

function getLength<[1]>(arr: [2][]): number {
  return arr.length;
}

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

AT

BU

DV

Attempts:

3 left

💡 Hint

Common Mistakes

Using different generic type names for the parameter and array element.

Not using a generic type for the array elements.

The generic type parameter is T, and the array parameter is of type T[].

5fill in blank

hard

Fill all three blanks to create a generic function that returns the first element of an array of generic type T or undefined if empty.

Typescript

function firstElement<[1]>(arr: [2][]): [3] | undefined {
  return arr.length > 0 ? arr[0] : undefined;
}

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

AT

BU

DV

Attempts:

3 left

💡 Hint

Common Mistakes

Using different generic type names for parameters and return type.

Not including undefined in the return type.

The generic type parameter is T, the array is of type T[], and the return type is T or undefined.

Practice

1. What does TypeScript do when you call a generic function without specifying the type explicitly?

easy

A. It ignores the generic and treats the function as non-generic.

B. It throws a compilation error asking for the type.

C. It guesses the generic type based on the arguments you pass.

D. It always uses the type any by default.

Solution

1. Step 1: Understand generic function calls

   When you call a generic function without specifying the type, TypeScript tries to figure out the type from the arguments you provide.

2. Step 2: Check TypeScript behavior

   TypeScript uses the argument types to infer the generic type automatically, so you don't have to write it explicitly.

3. Final Answer:

   It guesses the generic type based on the arguments you pass. -> Option C

4. Quick Check:

   Type inference = guessing type from arguments [OK]

Hint: Remember: TypeScript guesses generic types from arguments [OK]

Common Mistakes:

• Thinking TypeScript requires explicit generic types always
• Assuming it defaults to any type
• Believing generic types are ignored if not specified

2. Which of the following is the correct syntax to define a generic function in TypeScript that returns the same type it receives?

easy

A. function identity<T>(arg: T): T { return arg; }

B. function identity(arg: T): T { return arg; }

C. function identity<T>(arg): T { return arg; }

D. function identity<T>(arg: T): any { return arg; }

Solution

1. Step 1: Check generic function syntax

   The generic type parameter T must be declared before the function parameters inside angle brackets <T>.

2. Step 2: Verify parameter and return types

   The parameter arg should be typed as T, and the return type should also be T to return the same type.

3. Final Answer:

   function identity<T>(arg: T): T { return arg; } -> Option A

4. Quick Check:

   Generic syntax = <T> before parameters [OK]

Hint: Generic type <T> goes before parameters in function definition [OK]

Common Mistakes:

• Omitting <T> before parameters
• Not typing the argument as T
• Returning any instead of T

3. What will be the output of this TypeScript code?

function wrapInArray<T>(value: T) {
  return [value];
}

const result = wrapInArray(42);
console.log(typeof result[0]);

medium

A. "number"

B. "string"

C. "object"

D. "undefined"

Solution

1. Step 1: Identify the generic type inferred

   The function wrapInArray is called with 42, a number, so TypeScript infers T as number.

2. Step 2: Check the returned array and logged type

   The function returns an array containing the number 42. Accessing result[0] gives 42, so typeof result[0] is "number".

3. Final Answer:

   "number" -> Option A

4. Quick Check:

   typeof 42 = "number" [OK]

Hint: Generic type matches argument type, so typeof matches argument [OK]

Common Mistakes:

• Assuming the type is string by default
• Confusing array type with element type
• Expecting 'object' because of array

4. Consider this code snippet:

function firstElement<T>(arr: T[]): T {
  return arr[0];
}

const value = firstElement([true, false, true]);
console.log(value.toUpperCase());

What is the problem here?

medium

A. The array passed is empty, so arr[0] is undefined.

B. TypeScript cannot infer the generic type T.

C. The function firstElement returns an array instead of an element.

D. Calling toUpperCase() on a boolean causes a compile-time error.

Solution

1. Step 1: Analyze the inferred generic type

   The array [true, false, true] is of type boolean[], so T is inferred as boolean.

2. Step 2: Check the method call on the returned value

   The function returns a boolean, but toUpperCase() is a string method. Calling it on a boolean causes a compile-time error.

3. Final Answer:

   Calling toUpperCase() on a boolean causes a compile-time error. -> Option D

4. Quick Check:

   Boolean has no toUpperCase() method [OK]

Hint: Check if method matches the inferred type before calling [OK]

Common Mistakes:

• Thinking TypeScript inference failed
• Assuming function returns array instead of element
• Ignoring compile-time errors from wrong method calls

5. You want to write a generic function mergeObjects that takes two objects and returns a new object combining both. How does TypeScript infer the generic types when you call mergeObjects({name: "Alice"}, {age: 30})?

function mergeObjects(obj1: T, obj2: U): T & U {
  return {...obj1, ...obj2};
}

hard

A. TypeScript infers both T and U as object losing specific properties.

B. TypeScript infers T as {name: string} and U as {age: number}, returning their intersection type.

C. TypeScript cannot infer generic types for multiple parameters automatically.

D. TypeScript infers T and U as any by default.

Solution

1. Step 1: Identify argument types passed to the function

   The first argument is {name: "Alice"} which has type {name: string}. The second is {age: 30} with type {age: number}.

2. Step 2: Understand TypeScript's inference for multiple generics

   TypeScript infers T as the type of the first argument and U as the type of the second argument automatically.

3. Step 3: Check the return type

   The return type is the intersection T & U, meaning the combined object has both name and age properties.

4. Final Answer:

   TypeScript infers T as {name: string} and U as {age: number}, returning their intersection type. -> Option B

5. Quick Check:

   Multiple generics inferred from each argument [OK]

Hint: Each generic inferred from corresponding argument type [OK]

Common Mistakes:

• Assuming generic types default to object or any
• Thinking TypeScript can't infer multiple generics
• Confusing intersection type with union type