Why does this compile?

问题

I was taken aback earlier today when debugging some code to find that something like the following does not throw a compile-time exception:

 public Test () { 
     HashMap map = (HashMap) getList(); 
 }

 private List getList(){
     return new ArrayList();
 }

As you can imagine, a ClassCastException is thrown at runtime, but can someone explain why the casting of a List to a HashMap is considered legal at compile time?

回答1:

Because conceivably getList() could be returning a subclass of HashMap which also implements List. Unlikely, yes, but possible, and therefore compilable.

回答2:

For one thing List is an interface. There is no reason why there couldn't exist a subclass of HashMap which also implements the List interface. In this situation it would be perfectly valid.

来源：https://stackoverflow.com/questions/1495308/why-does-this-compile