typescript - 允许子类使用函数属性或方法覆盖

Question

class Foo {
  foo: () => void
}

class Bar extends Foo {
  foo() {}
}

有没有办法告诉 TypeScript 允许上面的例子？

Playground

Class 'Foo' defines instance member property 'foo', but extended class 'Bar' defines it as instance member function.

Answer 1

那是因为您可以调用父方法，但不能调用父实例成员。

原因是适当的方法被保存到原型(prototype)链中，而成员函数则没有。

class Foo {
    foo() {
        return `Successfully called Foo.foo()`;
    }

    bar = () => {
        return `Successfully called Foo.bar()`;
    }
}

console.log(
    ('foo' in Foo.prototype), // true (methods are correctly inherited)
    ('bar' in Foo.prototype), // false (instance properties are not inherited)
);

如果该属性不在属性链中，则尝试使用 super 调用它会导致运行时错误。

class Bar extends Foo {
    foo = () => {
        return super.foo(); // Good: `foo` is in the prototype chain
    }

    bar = () => {
        return super.bar(); // Runtime error: `bar` is not in the prototype chain
    }
}

这使得从方法转到类实例属性(这里:foo)是安全的，但反过来就不安全了。