Question

所以，我明白我可以在子类中使用super（）来调用在基类上定义的函数。但是，如果我想在其他地方调用该对象的超级方法，那就会发生轰炸

Parent.js

class Parent {
    yell() {
        console.log('yell')
    }
}

Child.js

class Child extends Parent {
     shout() {
         super.yell() //this works
     }
}


Child.super.yell() //this doesnt work

Answer 1

如果要在实例上调用super方法，请不要在子类中实现该方法（默认调用super方法）或调用{{ 1}}在子类方法实现中。

此外，您尝试在类本身而不是实例上调用方法，如果这是您的目标，则需要创建方法super.methodName()：

static

Answer 2

查看babel输出的已编译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);

这里有一些重要的事情：

您定义的方法将添加到类的原型
Child原型是Parent类的实例
调用super从原型中获取同名函数

所以为了叫喊你可以做以下几件事之一：

Object.getPrototypeOf(Object.getPrototypeOf(_Child)).yell.call(_Child)

或

Object.getPrototypeOf(Child.prototype).yell.call(_Child)

或者，我推荐这个：

Parent.prototype.yell.call(_Child)

ES6（Babel） - 不能在类定义之外调用扩展类的super.methodName

2 个答案: