Type of a generic method parameter

user_11906121 0
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Let's say I have a method with a generic parameter:

public <U> void genericMethod(U param) {
    // doStuff
}

I also have two classes, one of which extends the other:

public class A {}
public class B extends A {}

And I do this:

A var = new B();
genericMethod(var);

In genericMethod, what is the type of U?

The variable I passed genericMethod is declared as A, but is actually B, so is U of type A or B?

I've tried testing it but I can't find a way to get information on U at runtime.

inheritance generics java
3 Answers

Let's modify your example a little:

public class Example {
    static class A {
        int x;
    }
    static class B extends A {}

    static <U> U genericMethod(U param) {
        int i = param.x; // compile time error
        System.out.println(param.getClass()); // prints B
        return param;
    }

    public static void main(String... args) {
        A var1 = new B();
        A var2 = genericMethod(var1);
        B var3 = genericMethod(var1); // compile time error
    }
}

From the point of view of your invocation of genericMethod, U is of type A, so we can assign the return value to A, but not to B.

From the point of view of the implementation of genericMethod, U is of type Object, because we haven't put any bounds on it, so we can't, for instance, refer to a member field of A. If we wrote static <U extends A> U genericMethod(U param) then the type of U would be A.

The actual runtime type of param is B, as you can see if you call getClass()

user_110020

Generic types are resolved in compiler time. Therefore, <U> is determined by the compile-time type of var, which is A.

user_24227760

U is of type Object, because you can pass anything to genericMethod.
It could be a String, an Integer, an A or a B or anything else. The most common denominator of all objects is Object, so inside genericMethod the type of param will be Object.

If you need U to be an A or B, it could be defined as

public <U extends A> void genericMethod(U param) {

But again the only guarantee is it is (at least) an A and so you can access properties and methods of A, but not B (though B can override properties of A).

user_61519240
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