how to see the class name containing the actual code

Question

package a;
public class A {
  public String toString() {
    // return "I am an a.A"; is too primitive ;) 
    return "I am an " + getClass().getName(); // should give "a.A"
  }
}

--

package a;
public class X extends A {
  public static void main(String[] args) {
    X test = new X();
    System.out.println(test);  // returns "I am an a.X"
  }
}

I also tried with this.getClass() and super.getClass(). How can I get the class name of where toString() and getClass() is coded actually ? (a.A)

This is just a simplified sample, my point is how to avoid hard coding the base class name in the first file (A.java)

Already answered here : [link](http://stackoverflow.com/a/6271460/6296931) — Coder ACJHP, Nov 02 '16 at 11:50
@ Coder ACJHP: Not really, or I did not find it. The question there's different anyway. — datafiddler, Nov 02 '16 at 13:31

score 5 · Answer 1 · answered Nov 02 '16 at 11:50

5

package a;
public class A {
  public String toString() {
    return "I am an " + A.class.getName();
  }
}

should do the trick.

answered Nov 02 '16 at 11:50

glglgl

89,107
13
149
217

score 1 · Answer 2 · answered Nov 02 '16 at 12:13

1

just iterate over all super

public String toString() {
  Class cl = getClass();
  while (cl.getSuperclass() != Object.class) 
    cl = cl.getSuperclass();
  return cl.getName();
}

answered Nov 02 '16 at 12:13

rustot

331
1
11

Nyakiba · Accepted Answer · 2016-11-02T12:47:28.380

1

Change :

getClass().getName()

into

A.class.getName()

edited Nov 02 '16 at 12:47

answered Nov 02 '16 at 12:22

Nyakiba

862
8
18

`this.class` wont do - "A class literal is an expression consisting of the name of a class, interface, array, or primitive type followed by a `.' and the token class. It evaluates to an object of type Class, the class object for the named type (or for void). " – Gyro Gearless Nov 02 '16 at 12:40

how to see the class name containing the actual code

3 Answers3