我有两个打字稿课。一个扩展了另一个:
type NamedObject = {
name: string;
}
class AnyObjectManager {
objectList = [];
getAnyObject = (matches: (o: object) => boolean) => {
for (const o of this.objectList) {
if (matches(o)) {
return o;
}
}
return null;
}
}
class NamedObjectManager extends AnyObjectManager {
getObjectNamedTim = () => {
return this.getAnyObject(this.objectIsNamedTim)
}
objectIsNamedTim = (namedObject: NamedObject) => {
return namedObject.name === 'Tim';
}
}
在方法NamedObjectManager.getObjectNamedTim中,打字稿显示错误:
TS2345:类型'(namedObject:NamedObject)=>布尔值的参数不能分配给类型'(o:object)=>布尔值'的参数。参数“ namedObject”和“ o”的类型不兼容。类型“ {}”中缺少属性“名称”,但类型“ NamedObject”中则为必需。
现在,我知道NamedObjectManager将仅管理NamedObjects。假设像addObject这样的函数确保只有NamedObjects在这里。因此,如何告诉打字稿,由NamedObjectManager管理的所有对象都将是NamedObjects,并且我们可以在从父级继承的方法中安全地假定这一点?
您可以使用generics并以“类型安全”的方式管理对象:
class AnyObjectManager<T> {
objectList: T[] = [];
getAnyObject = (matches: (o: T) => boolean) => {
for (const o of this.objectList) {
if (matches(o)) {
return o;
}
}
return null;
}
}
class NamedObjectManager extends AnyObjectManager<NamedObject> {
getObjectNamedTim = () => {
return this.getAnyObject(this.objectIsNamedTim)
}
objectIsNamedTim = (namedObject: NamedObject) => {
return namedObject.name === 'Tim';
}
}
无关,getAnyObject
可以简化为:
getAnyObject = (matches: (o: T) => boolean) => this.objectList.find(matches) || null;