如何定义一个不能在TypeScript中返回任何内容的函数 [英] How to define a function which cannot return anything in TypeScript
问题描述
我正在编写一个具有2个非常相似的功能的API:
I am writing an API which has 2, very similar, functions:
function update(f: () => string) {...}
function updateDeep(f: () => void) {...}
如您所见,我正在尝试确保API的客户端根据调用的函数传递正确的函数类型.
As you can see, I am trying to make sure that the client of my API passes the correct function type depending on which function they call.
第一个函数update
按预期工作.正确地将引发编译错误:
The first function, update
, works as predicted. This will rightly throw a compilation error:
update(() => console.log('hey'));
第二个函数updateDeep
不会抛出编译错误事件,尽管它应该:
The second function, updateDeep
, does not throw a compilation error event though it should:
updateDeep(() => 'hey');
如何声明不返回任何内容的函数类型?
推荐答案
不可能做到这一点.作为函数的接收者,您唯一的能力就是为提供的函数设置一个 lower 下界.
It isn't possible to make this happen. As the recipient of a function, your only ability is to set a lower bound on what kind function is provided.
另请参阅TypeScript常见问题解答条目: https://github.com/Microsoft/TypeScript/wiki/FAQ#why-are-functions-returning-non-void-assignable-to-function-returning-void
See also the TypeScript FAQ entry: https://github.com/Microsoft/TypeScript/wiki/FAQ#why-are-functions-returning-non-void-assignable-to-function-returning-void
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