Typescript 为什么将诸如 Omit 之类的泛型应用于函数会导致它不可调用

我在玩TypeScript时发现，在可调用函数上应用Omit<T, K>会使其不再可调用:

declare function myCallableFunction(): void;

myCallableFunction(); // valid

type NonOmittedFunction = typeof myCallableFunction;
declare const myNonOmittedFunction: NonOmittedFunction;
myNonOmittedFunction(); // valid

type OmittedFunction = Omit<typeof myCallableFunction, 'foobar'>;
declare const myOmittedFunction: OmittedFunction;
myOmittedFunction(); // This expression is not callable.
                    //  Type 'OmittedFunction' has no call signatures.

为什么会这样？

以下是一个精心设计的示例，您可能希望在其中执行以下操作:

declare type CallCountingFunction = (() => void) & { count: number }
const myFunction: CallCountingFunction = (() => { 
                                            const x = () => {}; 
                                            x.count = 0; 
                                            return x;
                                         })()

myFunction.count; // valid
myFunction() // valid
type OmittedFunction = Omit<CallCountingFunction, 'count'>;
declare const myOmittedFunction: OmittedFunction;
myOmittedFunction(); // This expression is not callable.
                    //  Type 'OmittedFunction' has no call signatures.

Playground

对于涉及重新映射的all种通用实用程序类型(例如Partial<T>和Required<T>)，这似乎是正确的.