Typescript gives you the ability to create a versatile writing interface that includes all

Think about this:

interface Options {
    length?: number,
    width?: number
}

interface Action {
    performAction ({length, width}: Options): void
}

const myObject: Action = {
    performAction ({length, width}) {
    // do something without returning anything
    }
}

Both inputs are marked as optional, which means I can call performAction like this:

 myObject.performAction() // although it doesn't go through

I encountered an Error saying "Expected 1 argument, but received 0".

Am I overlooking anything? How should I define an Interface with all optional parameters?

Answer №1

It's important for your interface not to worry about default values or destructuring, as these are technical details. Instead, simply declare the parameter as optional, treating it like a scalar:

interface TSomeFunction {
    someFunction (size? : TArguments): void
}

The actual implementation can then handle both cases:

const someObject: TSomeFunction = {
    someFunction({ width, height } = {}) {
    // perform some operation, no need to return anything
    }
}

Answer №2

It seems like the error you are encountering may be originating from a different location. You could also try restarting your IDE as that often resolves frustrating issues :)

const someFunction = ({width, height}: TArguments = {}) => { ... }

For a correct implementation, refer to this TypeScript playground link

EDIT Remember to declare the parameter as optional, not just its keys:

interface TSomeFunction {
    someFunction (arguments?: TArguments): void; // Make sure to include a question mark after the parameter
}

// You can provide a default value for the function
const someObject: TSomeFunction = {
    someFunction ({width, height}: TArguments = {}) {}
}


// Alternatively, you can leave it optional without setting a default value
const someObject: TSomeFunction = {
    someFunction ({width, height}?: TArguments) {} // Don't forget the question mark
}

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