typescript 中是否有任何方法可以定义带有类型的可选字段？

Question

I need something like that:

interface A {
 a: SomeUtilityType<number>;
}

to be equivalent to:

interface A {
 a?: number;
}

I know that I can use Optional from 'utility-types':

import { Optional } from 'utility-types';

type B = Optional<A, 'a'>

but I need exactly usage like this:

interface A {
 a: SomeUtilityType<number>;
}

Answer 1

If you are looking how to define optional field in interface, you can use question mark the same as in you first example

interface A {
 a?: SomeUtilityType<number>;
}

?: operator can be used with primitives, and any other types too