Cannot access class methods via generics

Question

I changed my method to generic method. What is happening now is that I was deserializing the class inside the methodB and accessing its methods which I can not do anymore.

<T> void methodB(Class<T> clazz) {
    T var;
    HashMap<String, T> hash = new HashMap<>();
}

void methodA () {
  methodB(classA.class);
}

Initially inside methodB with no generics,

var = mapper.convertValue(iter.next(), ClassA.class);
var.blah() //works fine

After using generics,

var = mapper.convertValue(iter.next(), clazz);
var.blah() //cannot resolve the method.

How do I access blah() method of classA?

Error:() java: cannot find symbol. More code is in stackoverflow.com/questions/27413112/… — Ava
– Ava, Commented Dec 11, 2014 at 1:24
What's the type of var? What's the definition of that class? — Dima
– Dima, Commented Dec 11, 2014 at 1:31
What would happen if you called methodB(String.class)? String has no method blah. — Stack Exchange Broke The Law
– Stack Exchange Broke The Law, Commented Dec 11, 2014 at 1:39
This has nothing to do with serialization. It is a compiler error you are getting because your type parameter T is not restricted to a type having a blah method. You must specify a proper type boundary. — 5gon12eder
– 5gon12eder, Commented Dec 11, 2014 at 1:43

Don Vince · Accepted Answer · 2023-10-12 21:09:25Z

2

Thanks to Passing a class with type parameter as type parameter for generic method in Java. Solved using TypeToken

edited Oct 12, 2023 at 21:09

Don Vince

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answered Dec 11, 2014 at 8:50

Ava

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Comments

user2533521 · Accepted Answer · 2014-12-11 05:31:51Z

0

I think you should use interfaces instead of generics, if you want to call the same 'blah' function on a variety of classes (A,X,Y,Z) (each of which has the same function signature)..

Your other option (if you cannot modify A, e.t.c) is to use reflection. read more about this in https://docs.oracle.com/javase/tutorial/reflect/

answered Dec 11, 2014 at 5:31

user2533521

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Comments

newacct · Accepted Answer · 2014-12-12 02:09:57Z

0

The line where you assign to var at runtime is absolutely irrelevant. The only thing that matters for compiling a call is the static (compile-time) type of var. Since T is unbounded (other than by Object), it is not known to support any methods, other than those provided by Object. Both pieces of code should fail to compile.

answered Dec 12, 2014 at 2:09

newacct

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