Custom type guard functions help you check if a value is a specific type. This makes your code safer and easier to understand.
When you want to check if a variable is a certain object type before using it.
When you receive data from outside your program and need to confirm its shape.
When you want to narrow down union types to a specific type in your code.
When you want to avoid errors by making sure a value has the properties you expect.
Syntax
Typescript
function isTypeName(value: unknown): value is TypeName {
// return true if value matches TypeName
}
The function returns a boolean.
The return type uses value is TypeName to tell TypeScript about the type check.
Examples
This checks if a value is a string.
Typescript
function isString(value: unknown): value is string {
returntypeof value === 'string';
}
This checks if a value looks like a User object.
Typescript
interface User {
name: string;
age: number;
}
function isUser(value: unknown): value is User {
returntypeof value === 'object' && value !== null &&
'name'in value && typeof (value as any).name === 'string' &&
'age'in value && typeof (value as any).age === 'number';
}
Sample Program
This program defines two types: Cat and Dog. It uses a custom type guard isCat to check if a pet is a Cat. Then it calls the correct sound method.
Typescript
interface Cat {
name: string;
meow: () => void;
}
interface Dog {
name: string;
bark: () => void;
}
function isCat(pet: unknown): pet is Cat {
returntypeof pet === 'object' && pet !== null && 'meow'in pet;
}
function speak(pet: Cat | Dog) {
if (isCat(pet)) {
pet.meow();
} else {
pet.bark();
}
}
const myCat: Cat = { name: 'Whiskers', meow: () => console.log('Meow!') };
const myDog: Dog = { name: 'Rex', bark: () => console.log('Woof!') };
speak(myCat);
speak(myDog);
OutputSuccess
Important Notes
Custom type guards help TypeScript understand your checks better.
Always check for null when testing if a value is an object.
Use in operator to check if a property exists on an object.
Summary
Custom type guard functions check if a value is a specific type.
They return value is TypeName to help TypeScript narrow types.
Use them to make your code safer and clearer.
Practice
(1/5)
1. What is the main purpose of a custom type guard function in TypeScript?
easy
A. To declare a new type alias
B. To convert a value from one type to another
C. To check if a value matches a specific type and help TypeScript narrow types
D. To throw an error when a type mismatch occurs
Solution
Step 1: Understand what a type guard does
A type guard checks if a value is of a certain type at runtime.
Step 2: Recognize the benefit in TypeScript
It helps TypeScript narrow the type, making code safer and clearer.
Final Answer:
To check if a value matches a specific type and help TypeScript narrow types -> Option C
Quick Check:
Type guard = check type + narrow [OK]
Hint: Type guards confirm type and help TypeScript narrow it [OK]
Common Mistakes:
Thinking type guards convert types
Confusing type guards with error handling
Believing type guards declare new types
2. Which of the following is the correct syntax for a custom type guard function named isString that checks if a value is a string?
easy
A. function isString(value: any): value is string { return typeof value === 'string'; }
B. function isString(value: any): boolean { return typeof value === 'string'; }
C. function isString(value: any): string { return typeof value === 'string'; }
D. function isString(value: any): value is boolean { return typeof value === 'string'; }
Solution
Step 1: Identify the return type for a type guard
A custom type guard returns a type predicate like value is string.
Step 2: Check the function body
The function should return true if typeof value === 'string'.
Final Answer:
function isString(value: any): value is string { return typeof value === 'string'; } -> Option A
Quick Check:
Type guard syntax = value is Type [OK]
Hint: Type guard return type uses 'value is Type' syntax [OK]
Common Mistakes:
Using boolean instead of type predicate as return type
Returning wrong type predicate
Returning the type instead of a boolean
3. What will be the output of the following TypeScript code?
function isNumber(value: any): value is number {
return typeof value === 'number';
}
const input: any = 42;
if (isNumber(input)) {
console.log(input.toFixed(2));
} else {
console.log('Not a number');
}
medium
A. 42.00
B. Not a number
C. Error: toFixed is not a function
D. undefined
Solution
Step 1: Check the type guard function behavior
The function isNumber returns true if value is a number.
Step 2: Analyze the if condition and output
Since input is 42 (a number), the if block runs and input.toFixed(2) outputs '42.00'.
Final Answer:
42.00 -> Option A
Quick Check:
isNumber(42) = true, so prints 42.00 [OK]
Hint: If type guard true, use narrowed type safely [OK]
Common Mistakes:
Assuming input is not a number
Expecting runtime error on toFixed
Confusing output with string conversion
4. Identify the error in this custom type guard function and choose the fix:
function isArray(value: any): value is Array {
return Array.isArray(value);
}
medium
A. Change function name to 'isArrayType'
B. Change return type to 'value is any[]' instead of 'value is Array'
C. Add explicit type annotation to parameter 'value: unknown'
D. Replace 'Array.isArray' with 'typeof value === "array"'
Solution
Step 1: Understand the return type syntax
The return type should be a type predicate with a specific type, e.g., value is any[] for arrays.
Step 2: Identify the incorrect type usage
Array alone is a constructor, not a type for arrays. Use any[] or a typed array instead.
Final Answer:
Change return type to 'value is any[]' instead of 'value is Array' -> Option B
Quick Check:
Array type predicate = value is any[] [OK]
Hint: Use 'value is Type[]' for arrays in type guards [OK]
Common Mistakes:
Using 'Array' instead of 'any[]' or typed array
Changing function name unnecessarily
Using typeof 'array' which is invalid
5. Given the following types and function, which implementation correctly narrows the type to Dog using a custom type guard?
interface Dog { bark(): void; }
interface Cat { meow(): void; }
type Pet = Dog | Cat;
function isDog(pet: Pet): pet is Dog {
// Which implementation is correct?
}
hard
A. return pet.meow === undefined;
B. return pet instanceof Dog;
C. return typeof pet.bark === 'function';
D. return (pet as Dog).bark !== undefined;
Solution
Step 1: Understand the interfaces and type guard goal
We want to check if pet has a bark method to confirm it's a Dog.
Step 2: Evaluate the implementations
Directly accessing pet.bark or pet.meow causes TypeScript compile errors because those properties do not exist on the union type Pet. pet instanceof Dog does not work because Dog is an interface, not a class with a constructor. Using (pet as Dog).bark !== undefined compiles due to the type assertion and performs a runtime check to see if the bark property exists.
Final Answer:
return (pet as Dog).bark !== undefined; -> Option D
Quick Check:
Type assertion enables property check in union type guards [OK]
Hint: Use type assertion inside type guards for union property checks [OK]
Common Mistakes:
Using instanceof with interfaces
Directly accessing properties on union types
Checking absence of other type's method
Recommended
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