Call An Overridden Method From Super Class In Typescript
Answer : The key is calling the parent's method using super.methodName(); class A { // A protected method protected doStuff() { alert("Called from A"); } // Expose the protected method as a public function public callDoStuff() { this.doStuff(); } } class B extends A { // Override the protected method protected doStuff() { // If we want we can still explicitly call the initial method super.doStuff(); alert("Called from B"); } } var a = new A(); a.callDoStuff(); // Will only alert "Called from A" var b = new B() b.callDoStuff(); // Will alert "Called from A" then "Called from B" Try it here The order of execution is: A 's constructor B 's constructor The assignment occurs in B 's constructor after A 's constructor— _super —has been called: function B() { _super.apply(this, arguments); // MyvirtualMethod c...