What is the best way to make a generic type nullable?

Question

What is the best way to make a generic type nullable?

In the code snippet below, you can see a logging method:

private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    this.logger.log(operation + ' ' + this.url);
    if (requestData) {
        this.logger.log('SENT');
        this.logger.log(requestData);
    }
    this.logger.log('RECEIVED');
    this.logger.log(responseData);
    return responseData;
}

The requestData parameter is optional. The aim is to call logData without specifying the S type when requestData is not sent with the method. For instance, instead of using

this.logData<T, any>('GET', data)

, I would like to use

this.logData<T>('GET', data)

.

Is there a way to accomplish this?

generics typescript

Answer 1

Answer №1

Starting from TypeScript 2.3, the new feature of utilizing default values for generic parameters is available.

private logData<T, S = {}>(operation: string, responseData: T, requestData?: S) {
  // Add your custom implementation here
}

Answer 2

Starting from TypeScript 2.3, the new feature of utilizing default values for generic parameters is available.

private logData<T, S = {}>(operation: string, responseData: T, requestData?: S) {
  // Add your custom implementation here
}

Answer 3

Answer №2

New Update 2020: Using the keyword void will now render the generic type as optional.

type MyType<T = void> = AnotherType<T>;

The previous answer demonstrates how assigning a default value to an object makes it optional while still retaining its value.

Illustration with a default type value of {}:

type MainFunctionType<A, B> = (x:A, y:B) => void;

type CustomFunctionType<T = {}> = MainFunctionType<{title: string}, T>

const customFunction:CustomFunctionType = (x) => {

}

customFunction({ title: "John" });
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
// Expected 2 arguments, but received 1.(2554)

Code Playground Link Here

Demonstration with default generic type value set as void

type MainFunctionType<A, B> = (x:A, y:B) => void;

type CustomFunctionType<T = void> = MainFunctionType<{title: string}, T>

const customFunction:CustomFunctionType = (x) => {

}

customFunction({ title: "John" })

This resource provides valuable insights on implementing optional generics in Typescript.

Optional generic types usage guide

Access Playground Link

Answer 4

New Update 2020: Using the keyword void will now render the generic type as optional.

type MyType<T = void> = AnotherType<T>;

The previous answer demonstrates how assigning a default value to an object makes it optional while still retaining its value.

Illustration with a default type value of {}:

type MainFunctionType<A, B> = (x:A, y:B) => void;

type CustomFunctionType<T = {}> = MainFunctionType<{title: string}, T>

const customFunction:CustomFunctionType = (x) => {

}

customFunction({ title: "John" });
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
// Expected 2 arguments, but received 1.(2554)

Code Playground Link Here

Demonstration with default generic type value set as void

type MainFunctionType<A, B> = (x:A, y:B) => void;

type CustomFunctionType<T = void> = MainFunctionType<{title: string}, T>

const customFunction:CustomFunctionType = (x) => {

}

customFunction({ title: "John" })

This resource provides valuable insights on implementing optional generics in Typescript.

Optional generic types usage guide

Access Playground Link

Answer 5

Answer №3

In TypeScript 2.2, when you call this.logData("GET", data) with data of type T (you can experiment in the TS Playground), it will successfully infer as

this.logData<T, {}>("GET", data)

.

If the inference fails with your version of TypeScript, you can use the overload suggested by David Bohunek. Make sure to declare and define the second signature before using it for the available overloads.

// Declarations
private logData<T>(operation: string, responseData: T);
private logData<T, S>(operation: string, responseData: T, requestData?: S);
// Definition
private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    // Body
}

Answer 6

In TypeScript 2.2, when you call this.logData("GET", data) with data of type T (you can experiment in the TS Playground), it will successfully infer as

this.logData<T, {}>("GET", data)

.

If the inference fails with your version of TypeScript, you can use the overload suggested by David Bohunek. Make sure to declare and define the second signature before using it for the available overloads.

// Declarations
private logData<T>(operation: string, responseData: T);
private logData<T, S>(operation: string, responseData: T, requestData?: S);
// Definition
private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    // Body
}

Answer 7

Answer №4

If you are searching for how to define an optional generic type within a Type/Interface declaration, this explanation may be useful.

(I was in the same situation and struggled to find relevant information, but stumbled upon a great tip from John's solution which pointed me in the right direction.)

type MessageResponse = {
  message: string;
};

interface DataResponse<T> extends MessageResponse {
  data: T;
}

export type ObjectResponse<T = void> = T extends void
  ? MessageResponse
  : DataResponse<T>;

Answer 8

If you are searching for how to define an optional generic type within a Type/Interface declaration, this explanation may be useful.

(I was in the same situation and struggled to find relevant information, but stumbled upon a great tip from John's solution which pointed me in the right direction.)

type MessageResponse = {
  message: string;
};

interface DataResponse<T> extends MessageResponse {
  data: T;
}

export type ObjectResponse<T = void> = T extends void
  ? MessageResponse
  : DataResponse<T>;

Answer 9

Answer №5

What do you think of using the Partial utility type?

The Partial utility type creates a new type where all properties of a specified type are optional. This tool generates a type that includes all possible subsets of the original type.

Answer 10

What do you think of using the Partial utility type?

The Partial utility type creates a new type where all properties of a specified type are optional. This tool generates a type that includes all possible subsets of the original type.

Answer 11

Answer №6

void may not be suitable when expecting an object. However, it can work effectively if you take into account that void will overwrite any other type it is combined with (void & string essentially becomes just void) - this is likely the desired behavior.
unknown
{} is essentially equivalent to unknown

interface OffsetPagination {
    offset: number;
    limit: number;
}

interface PagePagination {
    page: number;
    size: number;
}

type HttpListParams<
    PaginationType extends OffsetPagination | PagePagination | void = void,
    Params = {
        filter?: string;
    },
> = PaginationType extends void ? Params : PaginationType & Params;

ts playground

Answer 12

void may not be suitable when expecting an object. However, it can work effectively if you take into account that void will overwrite any other type it is combined with (void & string essentially becomes just void) - this is likely the desired behavior.
unknown
{} is essentially equivalent to unknown

interface OffsetPagination {
    offset: number;
    limit: number;
}

interface PagePagination {
    page: number;
    size: number;
}

type HttpListParams<
    PaginationType extends OffsetPagination | PagePagination | void = void,
    Params = {
        filter?: string;
    },
> = PaginationType extends void ? Params : PaginationType & Params;

ts playground

Answer 13

Answer №7

If you want to implement the overloading method, here's an example:

private logData<T>(operation: string, responseData: T);
private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    this.logger.log(operation + ' ' + this.url);
    if (requestData) {
        this.logger.log('SENT');
        this.logger.log(requestData);
    }
    this.logger.log('RECEIVED');
    this.logger.log(responseData);
    return responseData;
}

However, it may not be necessary because instead of writing

this.logData<T, any>('GET', data)

, you can simply use this.logData('GET', data). The type T will be automatically inferred.

Answer 14

If you want to implement the overloading method, here's an example:

private logData<T>(operation: string, responseData: T);
private logData<T, S>(operation: string, responseData: T, requestData?: S) {
    this.logger.log(operation + ' ' + this.url);
    if (requestData) {
        this.logger.log('SENT');
        this.logger.log(requestData);
    }
    this.logger.log('RECEIVED');
    this.logger.log(responseData);
    return responseData;
}

However, it may not be necessary because instead of writing

this.logData<T, any>('GET', data)

, you can simply use this.logData('GET', data). The type T will be automatically inferred.

What is the best way to make a generic type nullable?

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

Answer №7

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