Javascript ES6(Babel)-无法在类定义之外调用扩展类的super.methodName
因此,我知道我可以在子类中使用super()调用在基类上定义的函数。然而,如果我想在别处调用该对象的超级方法,它会爆炸 Parent.jsJavascript ES6(Babel)-无法在类定义之外调用扩展类的super.methodName,javascript,inheritance,ecmascript-6,super,babeljs,Javascript,Inheritance,Ecmascript 6,Super,Babeljs,因此,我知道我可以在子类中使用super()调用在基类上定义的函数。然而,如果我想在别处调用该对象的超级方法,它会爆炸 Parent.js class Parent { yell() { console.log('yell') } } Child.js class Child extends Parent { shout() { super.yell() //this works } } Child.super.yell
class Parent {
yell() {
console.log('yell')
}
}
Child.js
class Child extends Parent {
shout() {
super.yell() //this works
}
}
Child.super.yell() //this doesnt work
如果要在实例上调用
super
方法,请不要在子类中实现该方法(默认情况下将调用super
方法),或者在子类方法实现中调用super.methodName()
此外,您试图对类本身而不是实例调用一个方法,如果这是您的目标,则需要使该方法成为静态的:
class Parent {
static yell() {
console.log('yell')
}
}
class Child extends Parent {
}
Child.yell();
看看babel输出的传输代码可能会有所帮助:
'use strict';
var _get = function get(_x, _x2, _x3) { var _again = true; _function: while (_again) { var object = _x, property = _x2, receiver = _x3; desc = parent = getter = undefined; _again = false; var desc = Object.getOwnPropertyDescriptor(object, property); if (desc === undefined) { var parent = Object.getPrototypeOf(object); if (parent === null) { return undefined; } else { _x = parent; _x2 = property; _x3 = receiver; _again = true; continue _function; } } else if ('value' in desc) { return desc.value; } else { var getter = desc.get; if (getter === undefined) { return undefined; } return getter.call(receiver); } } };
var _createClass = (function () { function defineProperties(target, props) { for (var i = 0; i < props.length; i++) { var descriptor = props[i]; descriptor.enumerable = descriptor.enumerable || false; descriptor.configurable = true; if ('value' in descriptor) descriptor.writable = true; Object.defineProperty(target, descriptor.key, descriptor); } } return function (Constructor, protoProps, staticProps) { if (protoProps) defineProperties(Constructor.prototype, protoProps); if (staticProps) defineProperties(Constructor, staticProps); return Constructor; }; })();
function _inherits(subClass, superClass) { if (typeof superClass !== 'function' && superClass !== null) { throw new TypeError('Super expression must either be null or a function, not ' + typeof superClass); } subClass.prototype = Object.create(superClass && superClass.prototype, { constructor: { value: subClass, enumerable: false, writable: true, configurable: true } }); if (superClass) subClass.__proto__ = superClass; }
function _classCallCheck(instance, Constructor) { if (!(instance instanceof Constructor)) { throw new TypeError('Cannot call a class as a function'); } }
var Parent = (function () {
function Parent() {
_classCallCheck(this, Parent);
}
_createClass(Parent, [{
key: 'yell',
value: function yell() {
console.log('yell');
}
}]);
return Parent;
})();
var Child = (function (_Parent) {
function Child() {
_classCallCheck(this, Child);
if (_Parent != null) {
_Parent.apply(this, arguments);
}
}
_inherits(Child, _Parent);
_createClass(Child, [{
key: 'shout',
value: function shout() {
_get(Object.getPrototypeOf(Child.prototype), 'yell', this).call(this);
}
}]);
return Child;
})(Parent);
或
或者,我推荐这个:
Parent.prototype.yell.call(_Child)
Child
不是Child
的实例,super
不是属性。你想要新的孩子().yell()
?哎呀,是的,让我来修一下。编辑:好的,是的,成功了。谢谢super
与非常类似。它不是实例的属性。从工程的角度来看,这也没有多大意义。调用方不应该对对象的继承链做出任何假设。它应该只关注它的界面。@minitech我想让你的答案成为公认的答案,但我不知道这是否可行,因为这是一个评论。@FelixKling Ahhh哇,是的,现在你这么说更有意义了。谢谢
Object.getPrototypeOf(Child.prototype).yell.call(_Child)
Parent.prototype.yell.call(_Child)