Typeguard is what you are looking for: function isDog(animal: Dog | Cat): animal is Dog { return "bark" in Dog }

Then: isDog(animal) ? animal.bark() : animal.meow() You get full type narrowing inside conditionals using typeguards.

You don't even need that. The code exactly as presented acts as a discriminator. TypeScript is smart enough to handle that logic in the if block and know whether animal has been validated as Dog vs Cat. GP is complaining about a feature that already exists in TypeScript

▲

littlecranky67 4 hours ago | parent [-]

It depends how you construct Dog and Cat. With Javascripts dynamic prototype chain, you could never know for sure.

▲

culi 4 hours ago | parent [-]

Try it

https://www.typescriptlang.org/play/?#code/C4TwDgpgBAIg9gcyg...

▲

reitzensteinm 3 hours ago | parent | next [-]

type Mutt = Dog & Cat

const imposter: Mutt = { bark: () => console.log("woof"), meow: () => console.log("meow"), }

You're both misunderstanding parent's point as well as the original point. Nobody ever claimed your link wouldn't compile.

	▲	culi 2 hours ago \| parent [-]
		I see what you mean, thanks Well imo GP is fundamentally misunderstanding TypeScript. It's explicitly a structural language not a nominal one. It goes against the entire design philosophy of TS

▲

4 hours ago | parent | prev [-]

[deleted]