What does the type parameter in the method definition mean?

Question

public class GenericMethods {
    public <T> void f(T x) {
        System.out.println(x.getClass().getName());
    }
    public static void main(String[] args) {
        GenericMethods gm = new GenericMethods();
        gm.f("");
        gm.f(1);
        gm.f(1.0);
        gm.f(1.0F);
        gm.f('c');
        gm.f(gm);
    }
}
/* Output:  
java.lang.String  
java.lang.Integer  
java.lang.Double  
java.lang.Float  
java.lang.Character  
GenericMethods  */

What does public <T> void f(T x) mean? Is it a return type?
But the function doesn't actually return anything. What does it refer to? I couldn't place it under any of the general function header categories : like acces-specifier, return-type,etc.

It's a generic type parameter, declaring that the method is generic, that's all. See https://docs.oracle.com/javase/tutorial/java/generics/methods.html — Jon Skeet, Oct 27 '16 at 08:33

score 1 · Accepted Answer · edited Sep 22 '17 at 10:45

1

It is a generic type parameter. See Oracle tutorial about Generics.

edited Sep 22 '17 at 10:45

Anatoly Shamov

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answered Oct 27 '16 at 08:34

Prabodh Mhalgi

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What does the type parameter in the method definition mean?

1 Answers1

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