Conditional Typing with Typescript

Question

I have two classes ClassA and ClassB as follow:

interface IClassA {
   functionA: ()=> void
}

class ClassA implements IClassA{
  functionA(){
    console.log('hello world A')
  }
}


interface IClassB {
   functionB: ()=> void
} 

class ClassB implements IClassB{
  functionB() {
    console.log('hello world B')
  }
}

I have another function that needs to takes an instance of ClassA or ClassB as parameter as shown below.

function sayHello(object) {
   // ...
}

How can I type the object in to access to the function functionA or functionB depending on the instance of the class being use? The code below won’t work:

function sayHello(object: IClassA | IClassB)

I don't want to use the generic any type. Thanks.

Soc · Accepted Answer · 2020-07-31 22:17:32Z

1

You can check for functionA in the object before trying to invoke it.

Example (playground):

function sayHello(obj: IClassA | IClassB) {
  if ('functionA' in obj) {
    obj.functionA()
  }
}

answered Jul 31, 2020 at 22:17

Soc

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Comments

John Pavek · Accepted Answer · 2020-07-31 23:52:21Z

1

Union Types

You can actually use the | operator to create a union type. If your classes have separate logic, you can use the instanceOf syntax to check your class before running class specific code.

function sayHello(input: ClassA | ClassB) : string {
  if (input instanceof ClassA) {
    input.A();
  }
}

edited Jul 31, 2020 at 23:52

answered Jul 31, 2020 at 21:58

John Pavek

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3 Comments

lomse Over a year ago

I'm getting Property 'functionA' does not exist on type 'ClassA | ClassB'. Property 'functionA' does not exist on type 'ClassB'

John Pavek Over a year ago

@lomse I've updated my answer as well. Please take a look.

lomse Over a year ago

Yeah this looks great. Thanks

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