Type inference with "in" operator

It's quite common in Typescript to have this kind of code:

interface Cat {
  meow(): void
}

interface Human {
  hello(): void
}

function talk(animal: Cat | Human) {
  if (/* it is a cat */) {
    animal.meow()
  } else {
    animal.hello()
  }
}

You can try something like this:

function talk(animal: Cat | Human) {
  if (typeof animal.meow === "function") {
    ...
  }
}

But won't work because you're trying to access "meow" before testing the type.

The standard solution is to use a typescript type guard function. Among other things, it's quite verbose.

I've just learned a better one: the javascript "in" operator:

function talk(animal: Cat | Human) {
  if ("meow" in animal) {
    animal.meow();
  } else {
    animal.hello();
  }
}

🖖