Constrain consistent discriminating union type usage

[alexeyMohnatkin]

Bug Report

When I use discriminating unions I expect that only one of them will be allowed by TS, not their intersection

🔎 Search Terms

Discriminating union

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about union types

⏯ Playground Link

Playground link with relevant code

💻 Code

type Common = {foo: string};
type A = Common & {a: string};
type B = Common & {b: string};
type Params = Common | A | B;

// expect error here
const bar: Params = {foo: '123', a: '123', b: '123'};

const myFn = (params: Params) => {}

// expect error here
myFn({foo: '123', a: '123', b: '123'})

🙁 Actual behavior

TS allows using both fields from A and B types

🙂 Expected behavior

TS should treat this as an error

Also if I try to set never for fields that I don't want to be set, it doesn't work

type A = Common & {a: string, b: never};
type B = Common & {a: never, b: string};

There's a workaround for this, but it looks hacky

type A = Common & {a: string, b?: never};
type B = Common & {a: never, b: string};

[MartinJohns]

Duplicate of #20863. And you also want #12936.

Additional properties is simply something you have to expect and deal with if you want to write robust code, given the current design of the language.

[alexeyMohnatkin]

@MartinJohns thank you.

For those who are looking for an answer:

type Common = {foo: string};
type A = {a: string, b?: never};
type B = {a?: never, b: string};
type C = {a?: never, b?: never};
type Params = Common & ( A | B | C);