I need to create a function that returns number if an optional value is provided, and number | null if not.

So far I've tried:

type GetValueFn = {
    (values: { defaultValue: number }): number | null;
    (values: { defaultValue: undefined }): number;
}

const getValue: GetValueFn = ({ defaultValue }) => {
    const randomValue = Math.ceil(Math.random() * 10)
    
    if (randomValue !== 5) return randomValue;
    if (defaultValue !== undefined) return defaultValue;
    return null;
}

But I get an error saying:

Type 'null' is not assignable to type 'number'

What's the correct way of doing this?

demo

CodePudding user response：

Yo can use only one definition.

? = undefined

type GetValueFn = (values: { defaultValue?: number }): number | null;

const getValue: GetValueFn = ({ defaultValue }) => {
    const randomValue = Math.ceil(Math.random() * 10)
    
    if (randomValue !== 5) return randomValue;
    if (!defaultValue) return defaultValue;
    return null;
}

CodePudding user response：

Apparently, function overloads in typescript still doesn't work correctly for arrow functions. more here. I don't know if this is still accurate. If you have the option to, you can declare the function like and it'll work correctly.

interface Values {
    defaultValue: number;
}

function getValue(values: Values) : number;
function getValue(values: Partial<Values>) : number | null;
function getValue(values: Values | Partial<Values>) {
    const randomValue = Math.ceil(Math.random() * 10)
    
    if (randomValue !== 5) return randomValue;
    if (values.defaultValue !== undefined) return values.defaultValue;
    return null;
}

Working in TS playground