Is there a way to declare c as an optional array of any type in this code snippet?
const a = ({ b, ...c }: { b: string, c: ? }) => null
Is there a way to declare c as an optional array of any type in this code snippet?
const a = ({ b, ...c }: { b: string, c: ? }) => null
Given that this involves property destructuring, the result will not be an array but rather an object:
const example = ({ b, ...c }: { b: string, c: object }) => null;
Here is a working example without TypeScript:
const example = ({ b, ...c }) => {
console.log("typeof c:", typeof c); // true
console.log("Array.isArray(c):", Array.isArray(c)); // false
console.log(JSON.stringify(c)); // '{"x":1,"y":2,"z":3}'
};
example({ x: 1, y: 2, z: 3 });Explained previously, a choice array of various kinds: c?: Array<any>
const exampleFunction = ({param1, param2}: { param1: string, param2?: Array<any> }): void => null
An illustration:
const exampleFunction = ({param1, param2}: { param1: string, param2?: Array<any> }): void => {
console.log('param1:', param1)
if (param2 instanceof Array) {
for (let item of param2) {
console.log('param2 item:', item)
}
}
}
exampleFunction({param1: 'value', param2: [1]})
exampleFunction({param1: 'value'})
{param1, param2}: involves unpacking, param1 and param2 are considered as variables.
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