Union types and extra properties

Question

Is there a way to make TypeScript compiler produce an error when a function is called with an argument that can be both of the union type cases? Example:

interface Name {
  name: string
}

interface Email {
  email: string
}

type NameOrEmail = Name | Email

function print(p: NameOrEmail) {
  console.log(p)
}

print({name: 'alice'}) // Works
print({email: 'alice'}) // Works
print({email: 'alice', name: 'alice'}) // Works, but I'd like it to fail
print({email: 'alice', age: 33}) // Doesn't work

Answer 1

You can use method overloading:

interface Name {
  name: string
}

interface Email {
  email: string
}

function print(p: Name): void;
function print(p: Email): void;
function print(p: Name | Email) {
  console.log(p);
}

print({name: 'alice'}) // Works
print({email: 'alice'}) // Works
print({email: 'alice', name: 'alice'}) // Doesn't work
print({email: 'alice', age: 33}) // Doesn't work

This will basically make the signature of the method implementation "invisible" to the rest of your code.

Demo

Edit:

As pointed out by @str in strict mode the overload signatures need to specify a return type. It's still possible for the implementation to have it's return type be inferred as long as it's compatible to the return types specified in the visible signatures.