Union type syntax and behavior in Typescript - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this union type check?

Consider the following TypeScript code. What will be logged to the console?

Typescript

type Result = string | number;

function printResult(value: Result) {
  if (typeof value === 'string') {
    console.log(value.toUpperCase());
  } else {
    console.log(value + 10);
  }
}

printResult('hello');
printResult(5);

A HELLO and 5

B "hello" and 15

C TypeError at runtime

D "HELLO" and 15

💡 Hint

Check how the typeof operator distinguishes between string and number.

The function checks if the value is a string, then converts it to uppercase. Otherwise, it adds 10 to the number. So 'hello' becomes 'HELLO' and 5 becomes 15.

🧠 Conceptual

intermediate

Which union type allows both string and boolean values?

Which of the following TypeScript union types correctly allows a variable to hold either a string or a boolean?

A type MyType = string | boolean;

B type MyType = string & boolean;

C type MyType = string, boolean;

D type MyType = string || boolean;

💡 Hint

Union types use the pipe symbol | to combine types.

The pipe symbol | creates a union type allowing either type. The ampersand & creates an intersection type, which requires both types simultaneously, which is impossible for string and boolean.

🔧 Debug

advanced

Why does this union type assignment cause an error?

Examine the code below. Why does the assignment cause a TypeScript error?

Typescript

type ID = number | string;

let userId: ID = true;

A Because union types cannot include boolean values.

B Because 'true' is assignable to 'number | string' but needs explicit casting.

C Because 'true' is not assignable to type 'number | string'.

D Because 'userId' must be declared with 'const' instead of 'let'.

💡 Hint

Check the allowed types in the union and the assigned value type.

The union type ID allows only number or string. The boolean value 'true' is not assignable to either, so TypeScript raises an error.

❓ Predict Output

advanced

What is the output of this function with union types and type narrowing?

What will the following TypeScript code output when executed?

Typescript

type Input = string | number | boolean;

function describe(input: Input) {
  switch (typeof input) {
    case 'string':
      return `String: ${input.length}`;
    case 'number':
      return `Number: ${input * 2}`;
    case 'boolean':
      return `Boolean: ${input ? 'yes' : 'no'}`;
  }
}

console.log(describe('abc'));
console.log(describe(4));
console.log(describe(false));

A "String: abc", "Number: 4", "Boolean: false"

B "String: 3", "Number: 8", "Boolean: no"

C "String: 3", "Number: 4", "Boolean: no"

D Runtime error due to missing default case

💡 Hint

Look at how typeof narrows the type and what each case returns.

For 'abc', length is 3. For 4, doubling gives 8. For false, ternary returns 'no'. The switch covers all types, so no runtime error.

🧠 Conceptual

expert

How does TypeScript handle union types with overlapping properties?

Given these interfaces and union type, what is the type of 'value.name' inside the function?

Typescript

interface A { name: string; age: number; }
interface B { name: string; location: string; }

type AB = A | B;

function printName(value: AB) {
  return value.name;
}

A string

B string | undefined

C never

D TypeScript error: Property 'name' does not exist on type 'AB'

💡 Hint

Check which properties are common to both interfaces in the union.

Both interfaces A and B have a 'name' property of type string. So in the union type AB, 'name' is guaranteed to be a string.

Practice

1. What does a union type in TypeScript allow a variable to do?

easy

A. Hold values of multiple specified types safely

B. Hold only one fixed type value

C. Automatically convert types at runtime

D. Restrict values to a single primitive type

Solution

1. Step 1: Understand union type purpose

   A union type lets a variable accept values from multiple specified types, not just one.

2. Step 2: Compare options with definition

   Only Hold values of multiple specified types safely correctly describes this behavior; others describe fixed or automatic conversions which union types do not do.

3. Final Answer:

   Hold values of multiple specified types safely -> Option A

4. Quick Check:

   Union type = multiple types allowed [OK]

Hint: Union types use | to allow multiple types [OK]

Common Mistakes:

• Thinking union types convert values automatically
• Believing union types restrict to one type only
• Confusing union with intersection types

2. Which of the following is the correct syntax to declare a variable value that can be either a string or a number?

easy

A. let value: string | number;

B. let value: string & number;

C. let value: string, number;

D. let value: string | number[];

Solution

1. Step 1: Recall union type syntax

   Union types use the pipe symbol | between types without parentheses needed.

2. Step 2: Evaluate each option

   let value: string | number; uses correct syntax: string | number. let value: string & number; uses & which means intersection, not union. let value: string, number; uses comma which is invalid. let value: string | number[]; uses number[] which is an array of numbers, not a single number.

3. Final Answer:

   let value: string | number; -> Option A

4. Quick Check:

   Union syntax uses | without commas [OK]

Hint: Use | between types, no commas or & for union [OK]

Common Mistakes:

• Using & instead of | for union types
• Separating types with commas
• Using number[] instead of number

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

function printId(id: string | number) {
  if (typeof id === 'string') {
    console.log('ID is a string: ' + id.toUpperCase());
  } else {
    console.log('ID is a number: ' + id.toFixed(2));
  }
}
printId(42);

medium

A. ID is a string: 42

B. ID is a number: 42

C. ID is a number: 42.00

D. Runtime error due to wrong method

Solution

1. Step 1: Analyze type check and method calls

   The function checks if id is a string or number. For number, it calls toFixed(2) which formats number with 2 decimals.

2. Step 2: Apply input and output

   Input is 42 (number), so else branch runs and prints 'ID is a number: 42.00'.

3. Final Answer:

   ID is a number: 42.00 -> Option C

4. Quick Check:

   Number uses toFixed(2) = 42.00 [OK]

Hint: Check typeof before calling type-specific methods [OK]

Common Mistakes:

• Calling string methods on numbers without check
• Expecting no decimal places in toFixed output
• Ignoring type narrowing with typeof

4. Identify the error in this TypeScript code using union types:

function formatInput(input: string | number) {
  return input.toUpperCase();
}

medium

A. No error, code works fine

B. Error: toUpperCase() not available on number

C. Error: Missing return type annotation

D. Error: Union types cannot be used in functions

Solution

1. Step 1: Check method usage on union type

   Method toUpperCase() exists on string but not on number.

2. Step 2: Understand type safety with union

   Without checking type, calling toUpperCase() on input causes error because input might be number.

3. Final Answer:

   Error: toUpperCase() not available on number -> Option B

4. Quick Check:

   Call methods only after type check [OK]

Hint: Always check type before calling type-specific methods [OK]

Common Mistakes:

• Calling string methods on union without type check
• Assuming union types allow all methods
• Ignoring compiler errors about method availability

5. Given this TypeScript code, what will be the output?

type Result = string | null;
function process(value: Result) {
  return value?.toUpperCase() ?? 'No value';
}
console.log(process(null));
console.log(process('hello'));

hard

A. No value\nhello

B. null\nhello

C. Error: Cannot call toUpperCase on null

D. No value\nHELLO

Solution

1. Step 1: Understand optional chaining and nullish coalescing

   The expression value?.toUpperCase() calls toUpperCase() only if value is not null or undefined. Otherwise, it returns undefined.

2. Step 2: Apply nullish coalescing fallback

   If the left side is undefined, ?? 'No value' returns 'No value'. So for null, output is 'No value'. For 'hello', it returns 'HELLO'.

3. Final Answer:

   No value HELLO -> Option D

4. Quick Check:

   Optional chaining + ?? handles null safely [OK]

Hint: Use ?. and ?? to safely handle null in unions [OK]

Common Mistakes:

• Expecting error when calling method on null
• Ignoring optional chaining usage
• Confusing null and undefined handling