打字稿中的默认泛型类型派生

Question

Why does "value.type" throw an error "Property 'type' does not exist on type 'P'"

class A<T extends {}, P = { type: keyof T, value: T[keyof T] }> {
  constructor (value: P) {
    if (value.type) {

    }
  }
}

Using extends is OK：

class A<T extends {}, P extends { type: keyof T, value: T[keyof T] }> {
  constructor (value: P) {
    if (value.type) {

    }
  }
}

Answer 1

A generic type parameter default (case 1) is different from a generic constraint (case 2).

Case 1: type parameter default

class A<P = { type: string, value: unknown }> {
  constructor(value: P) {
    if (value.type) { } // error, 'type' does not exist on type 'P' (good!)
  }
}

new A({ foo: "bar" })  // P is { foo: string; }, cannot rely on 'type' to be present.

You cannot rely on P to have a type property here. A default is only chosen, if the type parameter is not set by the caller (cannot be inferred and is not set manually). But that depends on the caller side.

Case 2: generic constraint

class AExtends<P extends { type: string, value: unknown }> {
  constructor(value: P) {
    if (value.type) { } // works
  }
}

new AExtends({ value: "bar" }) // error, 'type' is missing
new AExtends({ type: "foo", value: "bar" }) // works

The compiler can be sure, that value: P always has a property type . It is enforced by the constraint extends { type: keyof T, value: T[keyof T] } .

Code sample (simplified your code a bit; eg T is not needed)