Does TypeScript support default implementation of functions in interfaces [duplicate]

Question

Default implementation of functions that you want in the TypeScript interface

TypeScript Version：5.5.3 Node Version :18.14.2

interface Greeting {
    sayGoodbye(name: string): void = {
      console.log(`Goodbye, ${name}!`);
    };
  }
  class EnglishGreeting implements Greeting {
    sayGoodbye(name: string): void {
        throw new Error("Method not implemented.");
    }
    sayHello(name: string): void {
      console.log(`Hello, ${name}!`);
    }
  }

If the compilation fails in the above way, whether the interface can support the default implementation of the function

No, that's not possible. TS's interfaces are erased, they don't exist at runtime at all. See answers to the linked questions above for more information. You're presumably going to need to use class inheritance (possibly via mixins) if you want behavior like this. — jcalz
– jcalz, Commented Sep 14, 2024 at 3:10

Yash Raj Bharti · Accepted Answer · 2024-09-14 02:56:39Z

0

Well, the Interfaces must define method signatures without implementations. Meaning sayGoodbye(name: string): void = {...} is invalid Method like sayGoodbye(name: string): void is valid.


    interface Greeting {
      sayGoodbye(name: string): void;
    }
    
    class EnglishGreeting implements Greeting {
      sayGoodbye(name: string): void {
        console.log(`Goodbye, ${name}!`);
      }
    
      sayHello(name: string): void {
        console.log(`Hello, ${name}!`);
      }
    }

answered Sep 14, 2024 at 2:56

Yash Raj Bharti

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