must call super constructor
Uncaught ReferenceError: Must call super constructor in derived class before accessing 'this' or returning from derived constructor,So if we're making a constructor of our own, then we must call super , because otherwise the object with this reference to it won't be created. And we'll get an ... , 我父类和子类都试过返回一个对象，但是在子类中不调用 super() 依然会报错 Must call super constructor in derived class before accessing 'this' or ...,If the superclass has no no-arg constructor or it isn't accessible then not specifying the superclass constructor to be called (in the subclass constructor) is a compiler error so it must be specified. This is fine because if you add no constructor ex,Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s). ,However even if we are not using this we need a super() inside a constructor because ES6 class constructors MUST call super if they are subclasses . Thus, you ... , super([arguments]); // calls the parent constructor. super. ... When used in a constructor, the super keyword appears alone and must be used ...,,Yes, that sounds correct, albeit a bit oddly formulated. The rules should be. In a derived class, you always1 need to call the super(…) constructor; If you are not ... ,In a child class constructor, this cannot be used until super is called. ES6 class constructors MUST call super if they are subclasses, or they must explicitly return some object to take the place of the one that was not initialized.
| 相關軟體 SUPER Video Converter 資訊 | |
|---|---|
 must call super constructor 相關參考資料
Uncaught ReferenceError: Must call super constructor · Issue #24072 ...
Uncaught ReferenceError: Must call super constructor in derived class before accessing 'this' or returning from derived constructor https://github.com Class inheritance, super - The Modern Javascript Tutorial
So if we're making a constructor of our own, then we must call super , because otherwise the object with this reference to it won't be created. And we'll get an ... https://javascript.info ES6的class继承中，如果不想调用super(),则唯一的方法是让类的构造 ...
我父类和子类都试过返回一个对象，但是在子类中不调用 super() 依然会报错 Must call super constructor in derived class before accessing 'this' or ... https://segmentfault.com Is it unnecessary to put super() in constructor? - Stack Overflow
If the superclass has no no-arg constructor or it isn't accessible then not specifying the superclass constructor to be called (in the subclass constructor) is a compiler error so it must be speci... https://stackoverflow.com Why call super() in a constructor? - Stack Overflow
Moreover, if the parent's constructor takes parameters, and it has no default parameter-less constructor, you will need to call super() with argument(s). https://stackoverflow.com What does calling super() in a React constructor do? - Stack Overflow
However even if we are not using this we need a super() inside a constructor because ES6 class constructors MUST call super if they are subclasses . Thus, you ... https://stackoverflow.com super | MDN
super([arguments]); // calls the parent constructor. super. ... When used in a constructor, the super keyword appears alone and must be used ... https://developer.mozilla.org How to extend a class without having to using super in ES6 ...
https://stackoverflow.com When do I need to call `super` from a constructor? - Stack Overflow
Yes, that sounds correct, albeit a bit oddly formulated. The rules should be. In a derived class, you always1 need to call the super(…) constructor; If you are not ... https://stackoverflow.com How to extend a class without having to using super in ES6? - Stack ...
In a child class constructor, this cannot be used until super is called. ES6 class constructors MUST call super if they are subclasses, or they must explicitly return some object to take the place of ... https://stackoverflow.com |