Typescript 自动通用回调

我有以下代码片段:

// Library code

type Forced = {
    name: string;
}

const funcExample = <T extends Forced,>(
    values: T[],
    defaultValue: (name: string) => T,
): T[] => {
    // Do some operations...
    const result = values.map( (value) => {
        // Yay, I can access my .name attribute on a Generic.
        // That's what I want
        if (value.name === "wrong_condition_that_leads_to_default") {
            const res = defaultValue(value.name)
            return res;
        }
        const res2 = {...value}
        return res2;
    })

    return result;
}

// User code

type Animal = {
    name: string,
    parent: string,
}

const animals: Animal[] = [
    {name: "A", parent: "Y"},
    {name: "B", parent: "X"},
    {name: "C", parent: "Z"},
]

const myVar = funcExample(
    animals,
    (name) => {
        const t = {
            name: name,
            parent: "foo",
        };
        return t;
    }
)
// Type of myVar is...
// const myVar: {
//     name: string;
//     parent: string;
// }[]

// I want it to be...
// const myVar: Animal[]

我已经找到了一个解决方案，它包括调用我的泛型funcExample，并在回调中指定正确的类型，例如...

const myVar = funcExample(
    animals,
    (name) => {
        const t: Animal = {
            name: name,
            parent: "foo",
        };
        return t;
    }
)
// Now the type is great, automaticly.
// const myVar: Animal[]

有没有一种方法可以让它自动映射？

我今天想要解决的问题是，我想让TypeScrip使以下错误不可能发生:定义一个不尊重我在defaultValue: (name: string) => T类型签名中提供的签名的回调.

编辑:

例如，如果我没有实现键parent: string，这就不会抛出错误.但我希望它抛出一个错误.

const myVar = funcExample(
    animals,
    (name) => {
        const t = {
            name: name,
        };
        return t;
    }
)