正如其他人所提到的，get()等不是泛型的原因是，您正在检索的条目的键不必与传递给get()的对象的类型相同;该方法的规范只要求它们相等。这源于equals()方法如何接受Object作为参数，而不仅仅是对象的相同类型。

Although it may be commonly true that many classes have equals() defined so that its objects can only be equal to objects of its own class, there are many places in Java where this is not the case. For example, the specification for List.equals() says that two List objects are equal if they are both Lists and have the same contents, even if they are different implementations of List. So coming back to the example in this question, according to the specification of the method is possible to have a Map<ArrayList, Something> and for me to call get() with a LinkedList as argument, and it should retrieve the key which is a list with the same contents. This would not be possible if get() were generic and restricted its argument type.

原因是包含是由equals和hashCode决定的，它们是Object上的方法，并且都接受Object参数。这是Java标准库的早期设计缺陷。再加上Java类型系统中的限制，它强制依赖于equals和hashCode的任何东西接受Object。

在Java中获得类型安全哈希表和相等性的唯一方法是避免使用Object。equals和Object。hashCode并使用通用的替代品。函数式Java提供的类型类就是为了这个目的:Hash<A>和Equal<A>。提供了HashMap<K, V>的包装器，在其构造函数中接受Hash<K>和Equal<K>。因此，该类的get和contains方法采用类型为K的泛型参数。

例子:

HashMap<String, Integer> h =
  new HashMap<String, Integer>(Equal.stringEqual, Hash.stringHash);

h.add("one", 1);

h.get("one"); // All good

h.get(Integer.valueOf(1)); // Compiler error

正如其他人所提到的，get()等不是泛型的原因是，您正在检索的条目的键不必与传递给get()的对象的类型相同;该方法的规范只要求它们相等。这源于equals()方法如何接受Object作为参数，而不仅仅是对象的相同类型。

Although it may be commonly true that many classes have equals() defined so that its objects can only be equal to objects of its own class, there are many places in Java where this is not the case. For example, the specification for List.equals() says that two List objects are equal if they are both Lists and have the same contents, even if they are different implementations of List. So coming back to the example in this question, according to the specification of the method is possible to have a Map<ArrayList, Something> and for me to call get() with a LinkedList as argument, and it should retrieve the key which is a list with the same contents. This would not be possible if get() were generic and restricted its argument type.

还有一个更重要的原因，它在技术上不能做到，因为它破坏了地图。

Java has polymorphic generic construction like <? extends SomeClass>. Marked such reference can point to type signed with <AnySubclassOfSomeClass>. But polymorphic generic makes that reference readonly. The compiler allows you to use generic types only as returning type of method (like simple getters), but blocks using of methods where generic type is argument (like ordinary setters). It means if you write Map<? extends KeyType, ValueType>, the compiler does not allow you to call method get(<? extends KeyType>), and the map will be useless. The only solution is to make this method not generic: get(Object).

向后兼容，我想。Map(或HashMap)仍然需要支持get(Object)。

这是对波斯特尔定律的应用，“在你做的事情上要保守，在你接受别人的东西上要自由。”

无论类型如何，都可以执行相等性检查;equals方法定义在Object类上，并接受任何Object作为参数。因此，键等价和基于键等价的操作接受任何Object类型是有意义的。

当映射返回键值时，它通过使用类型参数尽可能多地保存类型信息。