Unknown type vs any type in Typescript - Practice Questions

Challenge - 5 Problems

❓ Predict Output

intermediate

Output difference between unknown and any types

What will be the output of the following TypeScript code?

Typescript

let a: unknown = 5;
let b: any = 5;

// Trying to add 10 to both variables
// console.log(a + 10); // Uncommenting this line causes error
console.log(b + 10);

A 15

B Compilation error on line with 'a + 10'

C TypeError at runtime

D NaN

💡 Hint

Consider how TypeScript treats operations on unknown vs any types.

The variable a is of type unknown, so TypeScript does not allow operations like addition without type checking. This causes a compilation error if you try a + 10. The variable b is of type any, so the addition is allowed and outputs 15.

🧠 Conceptual

intermediate

Why use unknown instead of any?

Which statement best explains why you might prefer unknown over any in TypeScript?

A <code>unknown</code> forces you to check the type before using the value, improving safety.

B <code>unknown</code> disables all type checking, making coding faster.

C <code>any</code> is safer because it prevents runtime errors.

D <code>any</code> requires explicit type assertions before use.

💡 Hint

Think about how each type affects type safety and error prevention.

unknown is safer because it forces the programmer to check the type before using the value, preventing many runtime errors. any disables type checking and can lead to bugs.

🔧 Debug

advanced

Identify the error with unknown type usage

What error will this TypeScript code produce?

Typescript

function process(value: unknown) {
  console.log(value.toFixed(2));
}
process(3.1415);

A Compilation error: Object is of type 'unknown'.

B Outputs '3.14' without error.

C Runtime error: toFixed is not a function.

D Compilation error: Missing return type.

💡 Hint

Consider what TypeScript allows you to do with unknown types without type assertions.

The value parameter is unknown, so you cannot directly call methods on it without first checking or asserting its type. This causes a compilation error.

❓ Predict Output

advanced

Result of type assertion from unknown to number

What will be the output of this TypeScript code?

Typescript

let val: unknown = "123";
let num = val as number;
console.log(num + 1);

A 124

B Runtime output: NaN

C Runtime output: '1231'

D Compilation error: Cannot assert string to number.

💡 Hint

Think about what happens when you assert a string as a number and then add 1.

The assertion val as number does not perform any runtime conversion; it only tells TypeScript to treat val as a number. At runtime, num remains the string "123", so num + 1 results in string concatenation: '1231'.

🧠 Conceptual

expert

Difference in type safety between unknown and any

Which statement correctly describes the difference in type safety between unknown and any in TypeScript?

A <code>any</code> requires type assertions, but <code>unknown</code> does not.

B <code>any</code> is safer because it allows flexible usage without errors, while <code>unknown</code> causes runtime errors.

C <code>unknown</code> and <code>any</code> behave identically in all cases.

D <code>unknown</code> is safer because it requires explicit type checks before usage, while <code>any</code> bypasses all checks.

💡 Hint

Focus on how each type affects compile-time safety and runtime behavior.

unknown forces the programmer to check or assert the type before using the value, preventing many errors. any disables type checking, allowing any operation, which can lead to runtime errors.