class MyClass<T> {
public void setValue(T value){
//insert code
}
public T getValue(){
return null;
}
}
那么我有一个具体的推导
class MyClassImp extends MyClass<String> {
@Override
public void setValue(String value){
//insert code
}
@Override
public String getValue(){
return null;
}
}
在MyClassImpl上反映为：
Class clazz = MyClassImpl.class; Method[] methods = clazz.getDeclaredMethods();
我得到两个超类实现java.lang.Object getValue(),void setValue(java.lang.Object)和java.lang.String getValue(),void setValue(java.lang.String).
根据Class.getDeclaredMethods()的Java文档
Returns an array of Method objects reflecting all the methods declared by the class or interface represented by this Class object. This includes public,protected,default (package) access,and private methods,but excludes inherited methods. The elements in the array returned are not sorted and are not in any particular order. This method returns an array of length 0 if the class or interface declares no methods,or if this Class object represents a primitive type,an array class,or void. The class initialization method
<clinit>is not included in the returned array. If the class declares multiple public member methods with the same parameter types,they are all included in the returned array.
为什么我得到超级类型的实现？
有没有我失踪的东西？
解决方法
基本上这是必需的,因为JVM真的不知道泛型(至少不是以深刻的方式) – 为了在JVM级别覆盖超类方法,它必须具有相同的签名.
看看javap -c MyclassImp的类,你会看到额外的合成方法：
public java.lang.Object getValue(); Code: 0: aload_0 1: invokevirtual #3; //Method getValue:()Ljava/lang/String; 4: areturn public void setValue(java.lang.Object); Code: 0: aload_0 1: aload_1 2: checkcast #4; //class java/lang/String 5: invokevirtual #5; //Method setValue:(Ljava/lang/String;)V 8: return }
