typescript 类似断言的类型保护

Question

这是否可能在没有 if 的情况下通过 never 返回的函数调用限制类型，例如 undefined like assert 在 typescript 中？

示例代码:

interface Foo { bar(): void }
function getFoo(): Foo | undefined { }

function test() {
    const foo = someService.getFoo();
    assert(foo);
    if (!foo) { // now mandatory because without this foo may be still undefined even if assert protects us from this
        return;
    }
    foo.bar(); // , here foo may be undefined
}

我希望能够以这样的方式编写 assert，这样我就可以跳过 if (!foo) 子句并拥有 foo类型限制为纯 Foo。

这在 Typescript 中可能吗？

我已经尝试使用 never 为抛出的类型添加重载:

function assertGuard(v: undefined | null | '' | 0 | false): never;
function assertGuard(v: any): void; // i'm not sure which one is  captured by TS typesystem here

function assertGuard<T>(v: T | undefined) {
    if (v === undefined || v === null || v === '' || v === 0 || v === false) {
         throw new AssertionError({message: 'foo'})
    }
}

这个可以编译，但是调用 assertGuard(foo) 无法识别 undefined 它将返回 never 所以不会将 foo 限制为 Foo。

我已经找到了可能的解决方法，但我认为经典的 assert 是一种更简洁的方法:

function assertResultDefined<T>(v: T|undefined): T | never {
    if (v === undefined) {
        throw new Error('foo');
    }
    return v;
}
function die(): never { throw new Error('value expected)}

const foo = assertResultDefined(getFoo()) // foo is Foo, undefined is erased
const foo = getFoo() || die();
    // undefined is erased from foo
    / CONS: doesn't play well with types that interpolate to `false` like 0, ''

Answer 1

Typescript 3.7 添加了 assertions in control flow analysis .

An asserts return type predicate indicates that the function returns only when the assertion holds and otherwise throws an exception

不再需要消费者方面的黑客攻击。

interface Foo { bar(): void }
declare function getFoo(): Foo | undefined;

function assert(value: unknown): asserts value {
    if (value === undefined) {
        throw new Error('value must be defined');
    }
}

function test() {
    const foo = getFoo();
    // foo is Foo | undefined here
    assert(foo);
    // foo narrowed to Foo
    foo.bar();
}

Playground

---

另外可以断言提供的参数是必需的类型:

declare function assertIsArrayOfStrings(obj: unknown): asserts obj is string[];

function foo(x: unknown) {
    assertIsArrayOfStrings(x);
    return x[0].length;  // x has type string[] here
}

Playground