When Omit<Type, Keys> breaks (my expectation)



I have scratched my head when Omit<Type, Keys> in TypeScript has not worked as I intended. This blog post will share with you what I have learned from reasoning why Omit has not worked.

What is Omit<Type, Keys>?

Omit<Types, Keys> is a utility type provided by TypeScript. It defines a new type that has all the properties of Type but omits properties specified to Keys. We can see it as a type operator that removes Keys from Type.

Suppose you have the following definitions,

interface IdentifiedPerson {
  id: number;
  name: string;
type Person = Omit<IdentifiedPerson, 'id'>;

The derived type Person will be equivalent to

interface Person {
  name: string;

When Omit<Type, Keys> breaks

Suppose I have the types similar to the following defined outside of my project.

interface IElement {
  [extensionName: string]: any;

interface InfoObject extends IElement {
  title: string;
  version: string;
  description?: string;

I want to make the title property of InfoObject optional. So I derive a new type from InfoObject by using Omit.

type InfoObjectOmittingTitle = Omit<InfoObject, 'title'> & {
  title?: string; // it is now optional

Now I can omit the title property to initialize an instance of InfoObjectOmittingTitle.

const info: InfoObject = {
  version: '0.0.1',
}; // ERROR! error TS2741: Property 'title' is missing in type '{ version: string; }' but required in type 'InfoObject'.

const info2: InfoObjectOmittingTitle = {
  version: '0.0.1',
}; // WORKS!

But a strange thing happens. The version property also has become optional contrary to my expectation.

const info3: InfoObjectOmittingTitle = {}; // WORKS!?

Additionally, even the following can be compiled!

const info4: InfoObjectOmittingTitle = {
  version: 123,
}; // WORKS!?

How Omit<Type, Keys> works

You can find the definition of Omit itself here in the source code of TypeScript. The following is an excerpt of the code.

type Omit<T, K extends keyof any> = Pick<T, Exclude<keyof T, K>>;

Omit is actually a composition of other utility types Pick and Exclude.

What is Pick<Type, Keys>?

Pick is the opposite of Omit. It defines a new type that has properties of Type included in Keys. The definition of Pick itself is here. The following is an excerpt of the code.

type Pick<T, K extends keyof T> = {
    [P in K]: T[P];

What is Exclude<UnionType, ExcludedMembers>?

Exclude defines a new union type that includes all the members of UnionType but excludes members in ExcludedMembers. The definition of Exclude itself is here. The following is an excerpt of the code.

type Exclude<T, U> = T extends U ? never : T;

What is going on in my failed attempt?

We are going to dissect the following part in the previous section.

Omit<InfoObject, 'title'>

This will be

Pick<InfoObject, Exclude<keyof InfoObject, 'title'>>

keyof InfoObject is equivalent to the following union type. Please refer to this page for how keyof works.

'title' | 'version' | 'description' | string;

string comes from the index signature [extensionName: string]: any of the base interface IElement.

So the Omit part will further become

Pick<InfoObject, Exclude<'title' | 'version' | 'description' | string, 'title'>>

The following table shows the evaluation of Exclude<T, 'title'> for each member T of 'title' | 'version' | 'description' | string. We have to understand "Distributive Conditional Types" to see what happens to the part Exclude<'title' | 'version' | 'description' | string, 'title'>.

TT extends 'title' ?Exclude<T, 'title'>

Thus the Exclude<'title' | 'version' | 'description' | string, 'title'> will be

'version' | 'description' | string

And the Omit part will become

Pick<InfoObject, 'version' | 'description' | string>

What happens to the final Pick? Here what I can do is only guessing from the consequence. I think the most inclusive member string would win at K extends keyof T or P in K during the expansion of Pick<T, K> and suppress other specific members 'version' and 'description'.

  [p: string]: InfoObject[string];

And the Pick would end up with

  [extensionsName: string]: any;

However, I have not confirmed any legitimate literature that backs up my guess yet.


According to this article, the following can be a workaround.

type InfoObjectWithOptionalTitle = Partial<InfoObject> & Pick<InfoObject, 'version'>;

Partial<Type> defines a new type that has every property of Type as optional. Intersection with Pick<InfoObject, 'version'> keeps the version property mandatory.

The major drawback is that it requires an exhaustive list of mandatory properties.


Omit<Type, Keys> may not work if Type contains an index signature. I have not confirmed in the TypeScript specification why this happens. But I have found a workaround.