Is this possible? Opposite of Partial Type

[opensrcken]

TypeScript Version: nightly

Code

interface Test {
 optional?: string;
}

function getProperty<K extends keyof Test>(t: Test, key: K): Test[K]|never {
 if (t[key]) {
 return t[key];
 }
 throw new Error('Value does not exist at key');
}

const a: string = getProperty({}, 'optional');

Target behavior:
The given function is coded such that it will always return a non-empty value, so it would be great if there were a way to make something like this compile. Is there a way to communicate the opposite of Partial<T>, e.g. that we know Test[K] will be non-empty?

Actual behavior:
error TS2322: Type 'string | undefined' is not assignable to type 'string'. Type 'undefined' is not assignable to type 'string'.

[mhegazy]

You can use inference from homomorphic mapped types in #12528 to get a type that removes the undefined

for instance:

interface Test {
 optional?: string;
 required: number;
}

type OptionalMap<T> = {
 [P in keyof T]: T[P] | undefined;
};

function getProperty<T, K extends keyof T>(t: T, key: K): T[K] {
 return t[key];
}

function getPropertyRemoveUndefined<T, K extends keyof T>(t: OptionalMap<T>, key: K): T[K] {
 // Put it in a value, since `t[p]` is not narrowed only t.p which is not possible here
 var value = t[key];
 if (value) {
 return value;
 }
 throw new Error('Value does not exist at key');
}

var t: Test = {required: 0};


var o1 = getProperty(t, 'optional'); // string | undefined
var r1 = getProperty(t, 'required'); // number

var o2 = getPropertyRemoveUndefined(t, 'optional'); // string
var r2 = getPropertyRemoveUndefined(t, 'required'); // number

[ahejlsberg]

I just merged #12589 and now you can simply use the predefined Partial<T> type:

interface Test {
 optional?: string;
 required: number;
}

function getProperty<T, K extends keyof T>(t: T, key: K): T[K] {
 return t[key];
}

function getPropertyRemoveUndefined<T, K extends keyof T>(t: Partial<T>, key: K): T[K] {
 // Put it in a value, since `t[p]` is not narrowed only t.p which is not possible here
 var value = t[key];
 if (value) {
 return value;
 }
 throw new Error('Value does not exist at key');
}

var t: Test = {required: 0};


var o1 = getProperty(t, 'optional'); // string | undefined
var r1 = getProperty(t, 'required'); // number

var o2 = getPropertyRemoveUndefined(t, 'optional'); // string
var r2 = getPropertyRemoveUndefined(t, 'required'); // number

[ahejlsberg]

opensrcken BTW, I noticed that you used a union Test[K] | never. Just wanted to point out that the never type is redundant here as it is always removed from union types (it effectively represents an empty union type).

[opensrcken]

It works. And point taken on never in unions. Thanks guys. 👍

[k8w]

So I must write a function to get these opposite type ?
Can I just remove undefined or null from a type just like use ! ?

[mjomble]

TypeScript 2.8 introduces Required<T> that may give you what you need: https://www.typescriptlang.org/docs/handbook/release-notes/typescript-2-8.html#example-6