How To Check If A Method Exists At Runtime In Java?

Question

How would one go about checking to see if a method exists for a class in Java? Would a try {...} catch {...} statement be good practice?

Answer 1

I assume that you want to check the method doSomething(String, Object).

You might try this:

boolean methodExists = false;
try {
  obj.doSomething("", null);
  methodExists = true;
} catch (NoSuchMethodError e) {
  // ignore
}

This will not work, since the method will be resolved at compile-time.

You really need to use reflection for it. And if you have access to the source code of the method you want to call, it's even better to create an interface with the method you want to call.

[Update] The additional information is: There is an interface that may exist in two versions, an old one (without the wanted method) and a new one (with the wanted method). Based on that, I suggest the following:

package so7058621;

import java.lang.reflect.Method;

public class NetherHelper {

  private static final Method getAllowedNether;
  static {
    Method m = null;
    try {
      m = World.class.getMethod("getAllowedNether");
    } catch (Exception e) {
      // doesn't matter
    }
    getAllowedNether = m;
  }

  /* Call this method instead from your code. */
  public static boolean getAllowedNether(World world) {
    if (getAllowedNether != null) {
      try {
        return ((Boolean) getAllowedNether.invoke(world)).booleanValue();
      } catch (Exception e) {
        // doesn't matter
      }
    }
    return false;
  }

  interface World {
    //boolean getAllowedNether();
  }

  public static void main(String[] args) {
    System.out.println(getAllowedNether(new World() {
      public boolean getAllowedNether() {
        return true;
      }
    }));
  }
}

This code tests whether the method getAllowedNether exists in the interface, so it doesn't matter whether the actual objects have the method or not.

If the method getAllowedNether must be called very often and you run into performance problems because of that, I will have to think of a more advanced answer. This one should be fine for now.