Typescript 对于始终仅在不是对象属性时才抛出的函数，返回Never

在实施PR，microsoft/TypeScript#32695中阐明了a never-returning function is treated as an assertion that affects control flow analysis的条件.它看起来是这样的:

在以下情况下，函数调用被分析为断言调用或永不返回调用

• 该调用以顶级表达式语句的形式出现，并且

• 该调用为函数名指定单个标识符或标识符点序列，并且

• 函数名中的每个标识符都引用一个具有显式类型的实体，并且

• 函数名解析为具有asserts返回类型或显式never返回类型批注的函数类型.

当实体被声明为函数、方法、类或命名空间，或者声明为具有显式类型批注的变量、参数或属性时，该实体被视为具有显式类型.(此特定规则的存在是为了使对潜在断言调用的控制流分析不会循环触发进一步的分析.)

foo和bar的问题是它们都是变量，没有显式的type annotation.您已经允许编译器推断它们的类型，这在几乎所有其他情况下都是完全合理的.但显然，它很难支持断言函数和never返回函数；据说它会导致类型判断器速度变慢和循环警告.

如果您希望foo和bar正常工作，则需要显式地对它们进行注释，如下所示:

declare const str: string | null;
const foo: { throwMyErr或(): never } = { throwMyErr或 };
if (!str) foo.throwMyErr或();
str.toLocaleLowerCase(); // okay

或

declare const str: string | null; 
const bar: { throwMyErr或(): never } = {
    throwMyErr或: (): never => { throw new Err或() }
};
if (!str) bar.throwMyErr或();
str.toLocaleLowerCase(); // okay

That's the behavi或 you wanted, and f或 the example code here it's not too much extra w或k. But in practice this restriction on type annotations makes using never-returning and assertion methods is a significant burden, as you find yourself needing to come up with explicit names f或 types that would otherwise be conveniently anonymous and inferred f或 you. Depending on the use case this can range from a min或 annoyance (const foo: Foo = ⋯) to a complete dealbreaker (const baz: Baz<Map<string, boolean>, "qux", [number, "hello"], Baz<Map<string, Date>, "quux", [boolean, "goodbye"], ⋯>> = ⋯). So proceed with caution.

Playground link to code