如何定义一个不能在TypeScript中返回任何内容的函数

Question

I am writing an API which has 2, very similar, functions:

function update(f: () => string) {...}
function updateDeep(f: () => void) {...}

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.

The first function, update , works as predicted. This will rightly throw a compilation error:

update(() => console.log('hey'));

The second function, updateDeep , does not throw a compilation error event though it should:

updateDeep(() => 'hey');

How to I declare a function type which does not return anything?

Answer 1

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.

See also the TypeScript FAQ entry: https://github.com/Microsoft/TypeScript/wiki/FAQ#why-are-functions-returning-non-void-assignable-to-function-returning-void