flag to check for type-unreachable code

feature request

the allowUnreachableCode flag (when disabled) only checks for code that's syntactically unreachable. however, the check doesn't work at all when looking at types.

i propose an additional flag (as @RyanCavanaugh suggested keeping these two concerns separate, see https://github.com/microsoft/TypeScript/issues/45267#issuecomment-901297945) that checks for code that the type checker believes is unreachable. for example:

declare function foo(): never

const a = foo()
console.log(1) //no error

foo()

console.log(1) //error

assertNever

in order to allow the assertNever pattern, the flag could still allow unreachable code where an argument of type never is passed, but not code after never is returned. this way, the following code would still be valid:

declare function assertNever(value: never): never
declare const foo: 1|2|3
switch(foo) {
    case 1:
        break
    case 2:
        break
    case 3:
        break
    default:
        assertNever(foo)
}

🔎 Search Terms

unreachable code

⏯ Playground Link

Playground link with relevant code

💻 Code

if (false) console.log("asdf") //error
if (false as false) console.log("asdf") //no error

リンクされている TypeScript Playground から始め、既存の allowUnreachableCode の動作を使って両方の例を再現します。これらのケースに関連する型チェックと到達不能コードの診断を読み、その後、never を返す場合と assertNever の呼び出しに対するフラグの動作を定義します。提案された型ベースの到達不能チェックと例外が、合意されたコンパイラテストでカバーされれば完了です。