Is it possible to define a type with an optional field in TypeScript?

I require a similar functionality:

interface A {
 a: SomeUtilityType<number>;
}

to be the same as:

interface A {
 a?: number;
}

I understand that I have the option to utilize Optional from 'utility-types':

import { Optional } from 'utility-types';

type B = Optional<A, 'a'>

However, my specific requirement is for it to be used in this manner:

interface A {
 a: SomeUtilityType<number>;
}

Answer №1

If you're wondering about defining an optional field in an interface, simply use a question mark as shown in the first example:

interface B {
 b?: AnotherUtilityType<string>;
}

The `?` operator is versatile and can be utilized with primitives as well as other types.

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