typescript - 正确处理 TypeScript 的弱类型检测

Question

鉴于这个简化的例子，表达该函数makeAnimalBark可以接受任何动物但只会让吠叫的动物吠叫的正确方法是什么？

interface Dog {
    bark: () => void
}

interface Cat {
    meow: () => void
}

interface PotentiallyBarkingAnimal {
    bark?: () => void
}

// Without `A` extending from something that has a `bark`
// property, TS complains that `bark` doesn't exist on `A`.

function makeAnimalBark<A extends PotentiallyBarkingAnimal>(animal: A) {
    if (animal && animal.bark)
        animal.bark()
}

makeAnimalBark<Dog>({
    bark: () => { }
})

// Here, TS complains that `Cat` has no properties
// in common with `PotentiallyBarkingAnimal`.

makeAnimalBark<Cat>({
    meow: () => { }
})

我理解为什么我会看到错误，但我不确定处理这种情况的正确方法。

标签： typescript

Answer 1

// Without `A` extending from something that has a `bark`
// property, TS complains that `bark` doesn't exist on `A`.

那是因为您输入bark的内容是可选的

interface PotentiallyBarkingAnimal {
    // remove the ? to make bark mandatory
    bark?: () => void
}

// Here, TS complains that `Cat` has no properties
// in common with `PotentiallyBarkingAnimal`.

那是真的，对吧？如果您查看它Cat的界面，它没有bark与PotentiallyBarkingAnimal. 您需要PotentiallyBarkingAnimal告知meow.