Typescript 泛型函数即使没有提供类型

我不明白为什么泛型函数自动获取泛型类型，而不提供任何东西.

type T<G = Record<string, any>> = {
     id: keyof G
     label: string
     somethingElse?: string
}

此函数返回一个具有特殊属性id的对象，如果没有将source属性传递给该对象，则该对象应该具有类型"random".如果传递source，则返回的id应为source的类型

const element = <ID extends string = 'random'>({ source }: { source?: ID } = {}) =>
    ({
        id: (source || 'random') as ID,
        label: 'Element1',
    }) as const satisfies T  
    // I don't want the return type to be "T", I want what it returns to satisfy "T".
    // as you see I dont provide "somethingElse" key here, so I don't want it in the return type

type Data = { a: string; b: number }

这给出了我想要的确切错误，但仅当我使用变量时:

const el1 = element()                   // el1 gets type { readonly id: "random";  readonly label: "Element1"; }
const el2 = element({source: 'b'})     // el2 gets type { readonly id: "b";  readonly label: "Element1"; }
const el3 = element({source: 'else'})   // el3 gets type { readonly id: "else";  readonly label: "Element1"; }

const b: T<Data>[] = [el, el2, el3]   // el1 and el3 invalid, el2 valid. It is all correct

但是为什么这是有效的，当我直接使用它没有变量？它应该抛出一个错误

const a: T<Data>[] = [element()]   
// No generic or property is provided to it "element()", 
// why the return type of "element()" is { readonly id: keyof Data; readonly label: "Element1"; } ?? 

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