Extends common class and implement interface - oop

Question

(This example written by Typescript but not only in Typescript case)
class IMyInterface {
doC:(any) => any;
}
class Common {
commonProperty:any;
doA() {
}
doB() {
}
}
class ClassA extends Common {}
class ClassB extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class Factory {
myClass: Common;
doSomething() {
// Property 'doC' does not exist on type 'Common'
this.myClass.doC('test');
}
}
Class A and B are extended Common class, so that in Factory class can define myClass type as Common.
But Class B need to implement IMyInterface, which Common class doesn't contained. So the Factory class throws an error that the interface method is not existed on Common class.
How and what's the best way to solve this?
[Edited]
First of all, #basarat thank you very much, but I'm still curious a little,
What if there are some more classes which implements IMyInterface
class ClassC extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class ClassD extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class ClassE extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
In that case, I can think, I can define the doC() method in Common class.
But also I want to make ClassB, C, D and E must implement the Doc method.
please advise me,

Answer 1

How and what's the best way to solve this
You basically want to say that myClass is generally just Common but in a special case it might be ClassB. You can do this using a union type + using a typeguard:
class Factory {
myClass: Common | ClassB;
doSomething() {
const myClass = this.myClass;
if (myClass instanceof ClassB){
// works!
myClass.doC('test');
}
}
}
More
Complete example:
class IMyInterface {
doC:(any) => any;
}
class Common {
commonProperty:any;
doA() {
}
doB() {
}
}
class ClassA extends Common {}
class ClassB extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class Factory {
myClass: Common | ClassB;
doSomething() {
const myClass = this.myClass;
if (myClass instanceof ClassB){
// works!
myClass.doC('test');
}
}
}
Docs
Union type : https://basarat.gitbooks.io/typescript/content/docs/types/type-system.html#union-type
Type Guard: https://basarat.gitbooks.io/typescript/content/docs/types/typeGuard.html
UPDATE
As requested, if one does want to test for the interface one needs to create a user defined type guard (docs https://basarat.gitbooks.io/typescript/content/docs/types/typeGuard.html#user-defined-type-guards). Example:
function isMyInterface(foo:any): foo is IMyInterface {
return typeof foo.doC === 'function';
}
class Factory {
myClass: Common | IMyInterface;
doSomething() {
const myClass = this.myClass;
if (isMyInterface(myClass)){
// works!
myClass.doC('test');
}
}
}
So complete code becomes :
class IMyInterface {
doC:(any) => any;
}
class Common {
commonProperty:any;
doA() {
}
doB() {
}
}
class ClassA extends Common {}
class ClassB extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class ClassC extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class ClassD extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
class ClassE extends Common implements IMyInterface {
doC(test:any) {
return true;
}
}
function isMyInterface(foo:any): foo is IMyInterface {
return typeof foo.doC === 'function';
}
class Factory {
myClass: Common | IMyInterface;
doSomething() {
const myClass = this.myClass;
if (isMyInterface(myClass)){
// works!
myClass.doC('test');
}
}
}