Typescript 如何在接口（如函数）中推断泛型值

Typescript有一个很好的特性，当给定类型时

type Foo = "foo"
type Bar = "bar"
type FooBar = Foo | Bar;
type IfFooThenBar<T extends FooBar> = T extends Foo ? Bar :
  T extends Bar ? Foo : never;

---

Typescript没有对变量的部分推断，您需要使用函数来推断接口的类型参数

要做到这一点，最短的方法就是生活：

type Foo = "foo"
type Bar = "bar"
type FooBar = Foo | Bar;
type IfFooThenBar<T extends FooBar> = T extends Foo ? Bar :
    T extends Bar ? Foo : never;



interface IIfFooThenBar<T extends FooBar> {
    fooOrBar: T
    barOrFoo: IfFooThenBar<T>
}

const a = (<T extends FooBar>(o: IIfFooThenBar<T>) => o)({
    fooOrBar: "foo",
    barOrFoo: "foo" // err
})

interface IIfFooThenBar<T extends FooBar> {
  fooOrBar: T
  barOrFoo: IfFooThenBar<T>
}

const a: IIfFooThenBar = {
  fooOrBar: "foo",
  barOrFoo: "foo" // this should result in an error
}

type Foo = "foo"
type Bar = "bar"
type FooBar = Foo | Bar;
type IfFooThenBar<T extends FooBar> = T extends Foo ? Bar :
    T extends Bar ? Foo : never;



interface IIfFooThenBar<T extends FooBar> {
    fooOrBar: T
    barOrFoo: IfFooThenBar<T>
}

const a = (<T extends FooBar>(o: IIfFooThenBar<T>) => o)({
    fooOrBar: "foo",
    barOrFoo: "foo" // err
})