TypeScript: How to Properly Type a Function that Accepts a Dynamic Object with Optional Properties?
I've searched everywhere and can't find a clear answer... I'm working on a TypeScript project where I need to create a function that takes an object with dynamic keys, but the properties of this object can be optional. I'm trying to ensure that the types are enforced correctly so that TypeScript can provide proper type checking. Here's what I have so far: ```typescript interface DynamicObject { [key: string]: string | number | undefined; } function processObject(obj: DynamicObject) { for (const key in obj) { if (obj.hasOwnProperty(key)) { const value = obj[key]; // Process the value console.log(`Key: ${key}, Value: ${value}`); } } } ``` However, when I try to call this function like this: ```typescript processObject({ a: 'Hello', b: 42, c: undefined }); ``` I get a warning that says `Type 'undefined' is not assignable to type 'string | number | undefined'`. This seems strange since I thought I handled the possibility of `undefined` in my interface. Additionally, I want to ensure that if a property is not present in the object, it should be treated as if it’s `undefined`. Is there a better way to type this function while also accommodating optional properties without losing the type safety? I'm using TypeScript version 4.5.2 and have tried adjusting the type definition to include optional properties using a `Partial` type, but I’m unsure if that’s the right approach. Any advice on how to resolve this issue or improve the type definition would be greatly appreciated! My development environment is Windows. I'm using Typescript 3.10 in this project. Any examples would be super helpful.