Is it possible to use optional destructured arguments in a Typescript function?

Question

Is it possible to use optional destructured arguments in a Typescript function?

Is there a way to create a function that accepts an optional object argument using destructuring in Typescript?

myFunction({opt1, opt2}?: {opt1?: boolean, opt2?: boolean})

The error message "A binding pattern parameter cannot be optional in an implementation signature" prevents this from working.

One workaround is to forego destructuring:

myFunction(options?: {opt1?: boolean, opt2?: boolean}) {
  const opt1 = options.opt1;
  const opt2 = options.opt1;
  ...

Despite their similarity, the two methods are not interchangeable.

The desire to use a destructured syntax stems from its convenience and natural flow. It also allows for concise default values:

myFunction({opt1, opt2 = true}?: {opt1?: boolean, opt2?: boolean})

If destructuring is not used, default values become obscured within the function implementation or require a more complex approach involving classes.

typescript

Answer 1

Answer №1

Consider using a default parameter:

function myCustomFunction({ param1, param2 = true }: { param1?: boolean; param2?: boolean } = {}) {
    console.log(param2);
}

myCustomFunction(); // Result: true

Using a default parameter prevents errors from destructuring undefined:

function myCustomFunction({ param1, param2 }) {
}
    
// Error: Cannot destructure property `param1` of 'undefined' or 'null'.
myCustomFunction();

Answer 2

Consider using a default parameter:

function myCustomFunction({ param1, param2 = true }: { param1?: boolean; param2?: boolean } = {}) {
    console.log(param2);
}

myCustomFunction(); // Result: true

Using a default parameter prevents errors from destructuring undefined:

function myCustomFunction({ param1, param2 }) {
}
    
// Error: Cannot destructure property `param1` of 'undefined' or 'null'.
myCustomFunction();

Answer 3

Answer №2

When no object is provided as an argument, destructuring cannot be done. In such cases, it's recommended to use a default object in the parameters like mentioned in the previous post:

type Options = { opt1?: boolean; opt2?: boolean; }

function myFunction({ opt1, opt2 }: Options = {}) {
    console.log(opt2, opt1);
}

myFunction() // undefined,  undefined 
myFunction({opt1: false}); // undefined,  false 
myFunction({opt2: true}); // true,  undefined

It's important to note that utilizing this destructuring pattern in the parameters is most beneficial under the following circumstances:

The number of options may vary
There could be potential changes in the function's API

Destructuring offers greater flexibility by allowing you to add multiple options with minimal impact on the function's API.

However, for simpler scenarios, a more basic approach can also be used:

// For functions that are not expected to change frequently:
function myFunctionBasic( opt1? :boolean, opt2?: boolean ) {
    console.log(opt2, opt1);
}

Answer 4

When no object is provided as an argument, destructuring cannot be done. In such cases, it's recommended to use a default object in the parameters like mentioned in the previous post:

type Options = { opt1?: boolean; opt2?: boolean; }

function myFunction({ opt1, opt2 }: Options = {}) {
    console.log(opt2, opt1);
}

myFunction() // undefined,  undefined 
myFunction({opt1: false}); // undefined,  false 
myFunction({opt2: true}); // true,  undefined

It's important to note that utilizing this destructuring pattern in the parameters is most beneficial under the following circumstances:

The number of options may vary
There could be potential changes in the function's API

Destructuring offers greater flexibility by allowing you to add multiple options with minimal impact on the function's API.

However, for simpler scenarios, a more basic approach can also be used:

// For functions that are not expected to change frequently:
function myFunctionBasic( opt1? :boolean, opt2?: boolean ) {
    console.log(opt2, opt1);
}

Answer 5

Answer №3

If you're looking to combine optional and required arguments while ensuring all values are non-null, this method provides the best approach:

export function createNewCar(properties: {
    year?: number
    make?: string
    model: string
    owner: [string, string]
}) {
    const defaults = {
        year: 1999,
        make: "toyota",
    }
    return { ...defaults, ...properties }
    // If you want a readonly Car:
    // return Object.freeze(properties)
}

export type Car = ReturnType<typeof createNewCar>;

const myCar = createNewCar({ model: "corolla", owner: ["John", "Doe"] });

export function displayCarDetails(car: Car) {
    console.log(`${car.owner}'s amazing ${car.year} ${car.model} made by ${car.make}`);
}

Answer 6

If you're looking to combine optional and required arguments while ensuring all values are non-null, this method provides the best approach:

export function createNewCar(properties: {
    year?: number
    make?: string
    model: string
    owner: [string, string]
}) {
    const defaults = {
        year: 1999,
        make: "toyota",
    }
    return { ...defaults, ...properties }
    // If you want a readonly Car:
    // return Object.freeze(properties)
}

export type Car = ReturnType<typeof createNewCar>;

const myCar = createNewCar({ model: "corolla", owner: ["John", "Doe"] });

export function displayCarDetails(car: Car) {
    console.log(`${car.owner}'s amazing ${car.year} ${car.model} made by ${car.make}`);
}

Is it possible to use optional destructured arguments in a Typescript function?

Answer №1

Answer №2

Answer №3

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