假设我有这样的代码：

const myGenericMethod = <T extends MyType1 | MyType2>(myList: T[]): T[] => {
  return myList; // simplified, my real method would return a shuffled list
};

type MyType1 = { name: 'Type1' };
type MyType2 = { name: 'Type2' };

const myList: MyType1[] | MyType2[] = [];
myGenericMethod(myList); // error!

最后一行将导致类型错误：

TS2345: Argument of type 'MyType1[] | MyType2[]' is not assignable to parameter of type 'MyType1[]'.
   Type 'MyType2[]' is not assignable to type 'MyType1[]'.
     Type 'MyType2' is not assignable to type 'MyType1'.
       Types of property 'name' are incompatible.
         Type '"Type2"' is not assignable to type '"Type1"'.

当我将myList创建更改为

const myList: MyType1[] | MyType2[] = [{ name: 'Type1' }];

但它会起作用。
我的TypeScript版本是4.9.4。我如何正确支持空列表？