Is it possible to assign a name to an implied return type?

Question

Is it possible to assign a name to an implied return type?

Imagine having a function like this:

function fetchUser() {
  return {
    username: "johndoe",
    email: "johndoe@example.com"
  }
}

You can create a type based on the return value:

type User = ReturnType<typeof fetchUser>;

But, when you hover over the function call or user below:

const user = fetchUser();

they are automatically given the type:

{
  username: string,
  email: string
}

However, it would be more beneficial if they were typed as User. Sometimes, it's important to know that an object was returned by a specific function. This is because functions may only return certain properties of the inferred types, and the type system may not be able to specify constraints like non-negative age.

So, ideally we could specify the User type in a way that when applying it to a function like:

function handleUser(user: User) {
  // ...
}

The following would be accepted:

handleUser(fetchUser());

but this would not:

handleUser({
  username: "johndoe",
  email: "johndoe@example.com"
});

This would result in an error. It's uncertain if TypeScript supports this feature or if it aligns with its design principles, but it would be useful. Is there a way to achieve this?

I have attempted different methods, but they have led to circular definition errors. If this feature does not exist, I will consider making a feature request. I am hoping for a syntax like:

function fetchUser(): infer User {

or

type User = ReturnType<typeof user as const>;

typescript

Answer 1

Answer №1

What you're really searching for is a nominal type, where only values of the type named Student (or a type from the same type declaration) are permitted. However, TypeScript's type system follows a largely structural approach rather than a nominal one. If two object types share the same members, they are considered the same type:

interface Student {
    firstName: string;
    lastName: string;
    age: number;
}
const student: Student = {
    firstName: "abc", lastName: "def", age: 123
};

interface Master {
    firstName: string;
    lastName: string;
    age: number;
}
const master: Master = student; // okay
// The student has become the master!!

These distinctions in TypeScript are intentional, unlike in other popular typed languages. If you wish for a different behavior, it's advised to understand and adapt with the system rather than trying to go against it. If there's a need for two types to be distinct, try to represent that difference structurally. Maybe one should have a method that the other doesn't? Finding a structural solution might lead to better outcomes.

Nevertheless, there are instances where nominal types are desired. TypeScript has an ongoing issue at microsoft/TypeScript#202, and many are in favor of it. There are also some techniques available to simulate/emulate nominal types.

One method is to include a "brand" member as detailed in the TypeScript FAQ, which reduces the chances of unintentional matches with other types. However, conflicts can still occur.

For more robust control, you can utilize a class with a private member to achieve this. This creates an "un-copyable" brand; private and protected class members are compared nominally and not structurally. Implementing this approach will align with the desired behavior:

class Student {
    firstName = "John";
    lastName = "Doe";
    age = 18;
    private readonly __type = "Student"
}
function passStudent(student: Student) { }

const student = new Student();
passStudent(student); // okay
passStudent(new Student()); // okay
passStudent({
    firstName: "John",
    lastName: "Doe",
    age: 18,
    __type: "Student"
}); // error!

In this scenario, the passed object literal matches the structure of Student, but the compiler raises an error because __type is private in Student. Even another class with a private member is rejected:

class Impostor {
    firstName = "John";
    lastName = "Doe";
    age = 18;
    private readonly __type = "Student" // 😈
}
passStudent(new Impostor()); // error!

For further exploration, Playground link to code

Answer 2

What you're really searching for is a nominal type, where only values of the type named Student (or a type from the same type declaration) are permitted. However, TypeScript's type system follows a largely structural approach rather than a nominal one. If two object types share the same members, they are considered the same type:

interface Student {
    firstName: string;
    lastName: string;
    age: number;
}
const student: Student = {
    firstName: "abc", lastName: "def", age: 123
};

interface Master {
    firstName: string;
    lastName: string;
    age: number;
}
const master: Master = student; // okay
// The student has become the master!!

These distinctions in TypeScript are intentional, unlike in other popular typed languages. If you wish for a different behavior, it's advised to understand and adapt with the system rather than trying to go against it. If there's a need for two types to be distinct, try to represent that difference structurally. Maybe one should have a method that the other doesn't? Finding a structural solution might lead to better outcomes.

Nevertheless, there are instances where nominal types are desired. TypeScript has an ongoing issue at microsoft/TypeScript#202, and many are in favor of it. There are also some techniques available to simulate/emulate nominal types.

One method is to include a "brand" member as detailed in the TypeScript FAQ, which reduces the chances of unintentional matches with other types. However, conflicts can still occur.

For more robust control, you can utilize a class with a private member to achieve this. This creates an "un-copyable" brand; private and protected class members are compared nominally and not structurally. Implementing this approach will align with the desired behavior:

class Student {
    firstName = "John";
    lastName = "Doe";
    age = 18;
    private readonly __type = "Student"
}
function passStudent(student: Student) { }

const student = new Student();
passStudent(student); // okay
passStudent(new Student()); // okay
passStudent({
    firstName: "John",
    lastName: "Doe",
    age: 18,
    __type: "Student"
}); // error!

In this scenario, the passed object literal matches the structure of Student, but the compiler raises an error because __type is private in Student. Even another class with a private member is rejected:

class Impostor {
    firstName = "John";
    lastName = "Doe";
    age = 18;
    private readonly __type = "Student" // 😈
}
passStudent(new Impostor()); // error!

For further exploration, Playground link to code

Answer 3

Answer №2

To ensure the correct type matching in your code, you can use a type assertion to explicitly define the return type of the function getStudent as Student. By doing this, any variable or function parameter defined as Student will only accept values that align with the structure of the Student type.

type Student = {
  firstName: string;
  lastName: string;
  age: number;
}

function getStudent(): Student {
  return {
    firstName: "Alice",
    lastName: "Smith",
    age: 22,
  };
}

function displayStudentInfo(student: Student) {
  console.log(student.firstName, student.lastName);
}

const studentInfo = getStudent() as Student;
displayStudentInfo(studentInfo);

Answer 4

To ensure the correct type matching in your code, you can use a type assertion to explicitly define the return type of the function getStudent as Student. By doing this, any variable or function parameter defined as Student will only accept values that align with the structure of the Student type.

type Student = {
  firstName: string;
  lastName: string;
  age: number;
}

function getStudent(): Student {
  return {
    firstName: "Alice",
    lastName: "Smith",
    age: 22,
  };
}

function displayStudentInfo(student: Student) {
  console.log(student.firstName, student.lastName);
}

const studentInfo = getStudent() as Student;
displayStudentInfo(studentInfo);

Answer 5

Answer №3

Implementing a wrapper function that seems to work:

function _getStudent() {
  return {
    firstName: "John",
    lastName: "Doe",
    age: 18,
  }
}

type Student = ReturnType<typeof _getStudent> & { __type: "Student" };

function getStudent(): Student {
  const student = _getStudent() as Student
  student.__type = "Student";
  return student;
}

see in playground

Below are the tests conducted:

const _student = _getStudent();
const student = getStudent();
const struct = {
  firstName: "John",
  lastName: "Doe",
  age: 18,
}
const typedStruct = {
  ...struct,
  __type: "Student"
}
const coercedStruct = typedStruct as Student
const propertyCoersion = {
  ...struct,
  __type: "Student" as "Student"
}

function passStudent(student: Student) {
  return true;
}

passStudent(_student);
passStudent(student);
passStudent(struct);
passStudent(typedStruct);
passStudent(coercedStruct);
passStudent(propertyCoersion);

see in playground

Although the tests above pass, it's important to note that the following scenario won't behave as expected:

passStudent({
  firstName: "John",
  lastName: "Doe",
  age: 18,
  __type: "Student",
});

As mentioned by jonrsharpe, the type inference doesn't work as intended in this case. However, combining the wrapper and private class approach as suggested by jcalz in the accepted answer can provide a solution with additional benefits:

This approach allows for encapsulation of validation logic without the need for bulky try-catch blocks:

class JWT {
  private __type = "JWT";
  public header: JWTHeader;
  public payload: JWTPayload;
  public signature: string;
 
  constructor(public token: string) {
    if(!isValidJWT(token)) throw "Invalid JWT";
    //...
  }

  //...

  toString() {
    return this.token;
  }
}

function parseJWT(input: string): JWT | null {
  try {
    return new JWT(input)
  } catch (error) {
    if(error === "Invalid JWT") return null;
    throw error;
  }
}

Answer 6

Implementing a wrapper function that seems to work:

function _getStudent() {
  return {
    firstName: "John",
    lastName: "Doe",
    age: 18,
  }
}

type Student = ReturnType<typeof _getStudent> & { __type: "Student" };

function getStudent(): Student {
  const student = _getStudent() as Student
  student.__type = "Student";
  return student;
}

see in playground

Below are the tests conducted:

const _student = _getStudent();
const student = getStudent();
const struct = {
  firstName: "John",
  lastName: "Doe",
  age: 18,
}
const typedStruct = {
  ...struct,
  __type: "Student"
}
const coercedStruct = typedStruct as Student
const propertyCoersion = {
  ...struct,
  __type: "Student" as "Student"
}

function passStudent(student: Student) {
  return true;
}

passStudent(_student);
passStudent(student);
passStudent(struct);
passStudent(typedStruct);
passStudent(coercedStruct);
passStudent(propertyCoersion);

see in playground

Although the tests above pass, it's important to note that the following scenario won't behave as expected:

passStudent({
  firstName: "John",
  lastName: "Doe",
  age: 18,
  __type: "Student",
});

As mentioned by jonrsharpe, the type inference doesn't work as intended in this case. However, combining the wrapper and private class approach as suggested by jcalz in the accepted answer can provide a solution with additional benefits:

This approach allows for encapsulation of validation logic without the need for bulky try-catch blocks:

class JWT {
  private __type = "JWT";
  public header: JWTHeader;
  public payload: JWTPayload;
  public signature: string;
 
  constructor(public token: string) {
    if(!isValidJWT(token)) throw "Invalid JWT";
    //...
  }

  //...

  toString() {
    return this.token;
  }
}

function parseJWT(input: string): JWT | null {
  try {
    return new JWT(input)
  } catch (error) {
    if(error === "Invalid JWT") return null;
    throw error;
  }
}

Is it possible to assign a name to an implied return type?

Answer №1

Answer №2

Answer №3

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