如何在typescript中实现回调返回类型的类型保护？

interface Bird {
    fly(): void;
}

interface Fish {
    swim(): void;
}

type Animal = Bird | Fish;

// 类型谓词函数
function isBird(animal: Animal): animal is Bird {
    return (animal as Bird).fly !== undefined;
}

// 使用类型谓词的回调函数
function doSomething(animal: Animal, callback: (isBird: boolean) => void): void {
    if (isBird(animal)) {
        // 在这个代码块中，TypeScript 知道 animal 是 Bird 类型
        animal.fly();
        callback(true);
    } else {
        // 在这个代码块中，TypeScript 知道 animal 是 Fish 类型
        animal.swim();
        callback(false);
    }
}

// 示例使用
const bird: Bird = { fly: () => console.log("Flying...") };
const fish: Fish = { swim: () => console.log("Swimming...") };

doSomething(bird, (isBird) => {
    console.log("Is bird?", isBird); // 输出: Is bird? true
});

doSomething(fish, (isBird) => {
    console.log("Is bird?", isBird); // 输出: Is bird? false
});