带有超类参数的Java getMethod方法

Question

Given:

class A
{
    public void m(List l) { ... }
}

Let's say I want to invoke method m with reflection, passing an ArrayList as the parameter to m :

List myList = new ArrayList();
A a = new A();
Method method = A.class.getMethod("m", new Class[] { myList.getClass() });
method.invoke(a, Object[] { myList });

The getMethod on line 3 will throw NoSuchMethodException because the runtime type of myList is ArrayList, not List.

Is there a good generic way around this that doesn't require knowledge of class A's parameter types?

Answer 1

If you know the type is List , then use List.class as argument.

If you don't know the type in advance, imagine you have:

public void m(List l) {
 // all lists
}

public void m(ArrayList l) {
  // only array lists
}

Which method should the reflection invoke, if there is any automatic way?

If you want, you can use Class.getInterfaces() or Class.getSuperclass() but this is case-specific.

What you can do here is:

public void invoke(Object targetObject, Object[] parameters,
        String methodName) {
    for (Method method : targetObject.getClass().getMethods()) {
        if (!method.getName().equals(methodName)) {
            continue;
        }
        Class<?>[] parameterTypes = method.getParameterTypes();
        boolean matches = true;
        for (int i = 0; i < parameterTypes.length; i++) {
            if (!parameterTypes[i].isAssignableFrom(parameters[i]
                    .getClass())) {
                matches = false;
                break;
            }
        }
        if (matches) {
            // obtain a Class[] based on the passed arguments as Object[]
            method.invoke(targetObject, parametersClasses);
        }
    }
}

Answer 2

请参阅java.beans.Expression和java.beans.Statement。

Answer 3

Instead of myList.getClass() , why not just pass in List.class ? That is what your method is expecting.

Answer 4

I'm guessing you want getDeclaredMethods() . Here is an example . You can dig through the list of methods and pick the one you want by name. Whether or not this is robust or a good idea is another question.