Determine the implicit type of the assigned function, while also constraining the return type to be a subtype of a predefined

Question

Determine the implicit type of the assigned function, while also constraining the return type to be a subtype of a predefined

When writing multiple functions for server requests, I have encountered a dilemma with TypeScript. Each function must return a type that extends a specific predefined known type, but I also want TypeScript to infer the most accurate return type possible.

For instance, if all functions are required to return a string:

type myFuncsType = (...args: any) => string

Each function should adhere to this type constraint. However, when a function always returns a constant string:

const myFunc1 = () => "MyString" as const
// The inferred type is:
const myVal = myFunc1()
// typeof myVal = "MyString"

We specified that our functions should extend a predefined known type (myFuncsType), yet when assigning this type to the function, the general type takes precedence over the inferred accurate type - something I wish to avoid:

const myFunc1: myFuncsType = () => "MyString" as const
const myVal = myFunc1()
// typeof myVal = string

I attempted to solve this issue using generics, but they require specifying a predefined type and do not account for the return type during declaration.

How can I enforce the restriction of the return type to extend a predefined type while still returning the exact inferred return type from the declaration?

typescript typescript-generics

Answer 1

Answer №1

Given that MyFuncsType is not a union, if you provide a type annotation for the variable myFunc1 with that type, the compiler will always perceive that variable as belonging to that specific type. It does not narrow down the variable based on the value assigned to it; instead, it widens it to match the annotated type. Therefore, avoid annotating myFunc1.

Instead of using annotations, what you actually need to do is simply verify that myFunc1 can be assigned to MyFuncsType, without widening it. TypeScript currently lacks a built-in operator for this purpose; refer to microsoft/TypeScript#7481 for the request for such a feature. However, you can create your own utility function to achieve this behavior:

type MyFuncsType = (...args: any) => string
const asMyFuncsType = <T extends MyFuncsType>(t: T) => t;

Instead of writing const f: MyFuncsType = ..., use const f = asMyFuncsType(...) instead. The asMyFuncsType() function simply returns its input without altering its type, but it validates that the type can be assigned to MyFuncsType, thereby capturing potential errors:

const badFunc = asMyFuncsType(() => 123); // error!
// -------------------------------> ~~~
// Type 'number' is not assignable to type 'string'

const myFunc1 = asMyFuncsType(() => "MyString" as const); // okay
const myVal = myFunc1()
// typeof myVal = "MyString"

Link to code in Playground

Answer 2

Given that MyFuncsType is not a union, if you provide a type annotation for the variable myFunc1 with that type, the compiler will always perceive that variable as belonging to that specific type. It does not narrow down the variable based on the value assigned to it; instead, it widens it to match the annotated type. Therefore, avoid annotating myFunc1.

Instead of using annotations, what you actually need to do is simply verify that myFunc1 can be assigned to MyFuncsType, without widening it. TypeScript currently lacks a built-in operator for this purpose; refer to microsoft/TypeScript#7481 for the request for such a feature. However, you can create your own utility function to achieve this behavior:

type MyFuncsType = (...args: any) => string
const asMyFuncsType = <T extends MyFuncsType>(t: T) => t;

Instead of writing const f: MyFuncsType = ..., use const f = asMyFuncsType(...) instead. The asMyFuncsType() function simply returns its input without altering its type, but it validates that the type can be assigned to MyFuncsType, thereby capturing potential errors:

const badFunc = asMyFuncsType(() => 123); // error!
// -------------------------------> ~~~
// Type 'number' is not assignable to type 'string'

const myFunc1 = asMyFuncsType(() => "MyString" as const); // okay
const myVal = myFunc1()
// typeof myVal = "MyString"

Link to code in Playground

Answer 3

Answer №2

When declaring a function, we have the option to specify its return type.

An effective way to determine the return type of a function in TypeScript is by using the built-in utility ReturnType. This utility helps extract the return type from the function type. To explore more examples on how to use the ReturnType utility, you can visit this page.

type myFuncsType = (...args: any) => string
type FnReturnType = ReturnType<myFuncsType>;

Your query: How can I constrain the return type to be an extension of a predefined type while still returning the inferred return type?

Firstly, let's distinguish between an extended type (or assigned type) and a type assertion.

const a: any = 20;
const b = a as number;
// typeof b is number

By assigning a type to a variable, TypeScript determines and confirms the expected type for that variable.

We resort to type assertions when we possess specific knowledge about a value's type that TypeScript is unaware of.

Therefore, trying to assign both an assigned type and a type assertion for the same value would result in TypeScript asserting its superior knowledge of the type over yours.

Answer 4

When declaring a function, we have the option to specify its return type.

An effective way to determine the return type of a function in TypeScript is by using the built-in utility ReturnType. This utility helps extract the return type from the function type. To explore more examples on how to use the ReturnType utility, you can visit this page.

type myFuncsType = (...args: any) => string
type FnReturnType = ReturnType<myFuncsType>;

Your query: How can I constrain the return type to be an extension of a predefined type while still returning the inferred return type?

Firstly, let's distinguish between an extended type (or assigned type) and a type assertion.

const a: any = 20;
const b = a as number;
// typeof b is number

By assigning a type to a variable, TypeScript determines and confirms the expected type for that variable.

We resort to type assertions when we possess specific knowledge about a value's type that TypeScript is unaware of.

Therefore, trying to assign both an assigned type and a type assertion for the same value would result in TypeScript asserting its superior knowledge of the type over yours.

Determine the implicit type of the assigned function, while also constraining the return type to be a subtype of a predefined

Answer №1

Answer №2

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