typescript Equal函数相等性推测

李恒道

type Equal<X, Y, A = X, B = never> = (<T>() => T extends X ? 1 : 2) extends <
  T
>() => T extends Y ? 1 : 2
  ? A
  : B;

这里的

<T>() => T extends X ? 1 : 2) extends <
  T
>() => T extends Y ? 1 : 2

实际只是一个辅助作用
并没有参与实际推断
具体可以参考
https://github.com/Microsoft/TypeScript/issues/27024

@jituanlin AFAIK it relies on conditional types being deferred when T is not known. Assignability of deferred conditional types relies on an internal isTypeIdenticalTo check, which is only true for two conditional types if:

Both conditional types have the same constraint
The true and false branches of both conditions are the same type

当函数T是未知时，依赖于延迟推断类型，延迟推断类型的可分配性依赖于内部isTypeIdenticalTo的检查，仅在两种情况下为真，
1.extends分支结果是相同类型，
2.两个条件类型是同一类型，
即

<T>() => T extends X ? 1 : 2 和 <T>() => T extends X

时返回true，通过该方法判断类型是否是readonly
如

export type MutableKeys<T extends object> = {
  [P in keyof T]-?: Equal<
    { [Q in P]: T[P] },
    { -readonly [Q in P]: T[P] },
    P,
    never
  >;
}[keyof T];

expectType<MutableKeys<{ a: string; readonly b: string }>>('a');

但更具体的理论就不清楚了....