Using the "in" operator should make bracket access safe on an object

Search Terms

object in keyof bracket operator in

Suggestion

Checking a key with the “in” operator on an object should make it safe to to use in brackets

Use Cases

Useful when writing functions that process input JSON/doesn’t have a concrete type.

Examples

//  Ideally this uses TypeScripts control flow logic to allow the bracket access.
function getPropFailsToCompile(data: object, key: string): any {
    if (!(key in data)) {
        throw new Error("Data is malformed")
    }
    return data[key]
}

// Compiles but keyof object has type 'never' in Typescript
function getPropCompilesButNotReasonable(data: object, key: keyof object): any {
    if (!(key in data)) {
        throw new Error("Data is malformed")
    }
    return data[key]
}

// Best way I've gotten this to work.
function getPropWorks(data: object, key: string): any {
    if (!(key in data)) {
        throw new Error("Data is malformed")
    }
    return (<any>data)[key]
}

Checklist

My suggestion meets these guidelines:

• [x ] This wouldn’t be a breaking change in existing TypeScript/JavaScript code
• [x ] This wouldn’t change the runtime behavior of existing JavaScript code
• [x ] This could be implemented without emitting different JS based on the types of the expressions
• [x ] This isn’t a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, etc.)
• [ x] This feature would agree with the rest of TypeScript’s Design Goals.

[nightpool]

I had the same issue today (Playground Link)

I was able to get around it by defining an existsIn helper as follows:

function existsIn<T>(obj: T, name: string | number | symbol): name is keyof T {
  return name in obj;
}

And then I was able to define the function I wanted

const values = {test: 'this value', foo: 'that value'};

const handleMessage = (name: string): void => {
  if (existsIn(values, name)) { // instead of `name in values`
    console.log(values[name]);
  }
}

maybe there’s some unsoundness I haven’t considered here, but it seems like it should be pretty simple to have name in obj imply the same name is keyof T assertion that an explicit type guard does!

[stuft2]

@RyanCavanaugh What is the typescript recommended way to determine that an unknown variable is an object with a given property on it?

@nightpool’s suggestion is fairly good but like he said, “it should be pretty simple to have name in obj imply the type correctly”. I’m always typing something like the following example even thought it doesn’t work and I intuitively think it should:

const obj: unknown = { foo: 'bar'}
if (typeof obj === 'object' && obj !== null && Object.hasOwnProperty.call(obj, 'foo') && typeof obj.foo === 'string') {
  //                                                                                             ^^^^^ ERROR
}

A common use case is when catching an error. The error must either be typed as unknown or any but it’s a hassle to type check an error that has additional properties on it. For example, node errors usually have a code property on it that can’t be type checked without the ritual.

There needs to be a documented, typescript-recommended way of accomplishing this.