How to mock an interface default method Java8/Mockito2












1















I cannot mock a method defined as default in an interface. Can anyone help me here?



The interface has default method providing a logger.



import org.slf4j.Logger;

import org.slf4j.LoggerFactory;



public interface Loggable {



  default Logger logger() {

    return LoggerFactory.getLogger(this.getClass());

  }

}


It is used this way:



public class AppShowOff implements Loggable{



  public void doMagic() {

    logger().debug("It works");

    System.out.println("Works");

  }

}


now I would like to write a test proving that debug method has been called.



import static org.mockito.ArgumentMatchers.any;

import static org.mockito.Mockito.mock;

import static org.mockito.Mockito.times;

import static org.mockito.Mockito.verify;

import static org.mockito.Mockito.when;



import org.junit.Test;

import org.mockito.Mockito;

import org.slf4j.Logger;



public class AppShowOffTest {



  @Test

  public void doMagic() {

    Logger loggerMock = mock(Logger.class);

    Loggable loggableMock = mock(Loggable.class); // <- not needed, but I also tried this way



    // mocks done



    AppShowOff app = new AppShowOff();

    AppShowOff appSpy = Mockito.spy(new AppShowOff());



    when(loggableMock.logger()).thenReturn(loggerMock);

    when(appSpy.logger()).thenReturn(loggerMock);



    app.doMagic();

    verify(loggerMock, times(1)).debug(any());

  }

}


as you can see I have tried to mock the default method in two ways: 



when(loggableMock.logger()).thenReturn(loggerMock);

when(appSpy.logger()).thenReturn(loggerMock);


but it does not work. The result is:




Wanted but not invoked: logger.debug();
-> at so.AppShowOffTest.doMagic(AppShowOffTest.java:29) Actually, there were zero interactions with this mock.







java unit-testing mockito







share|improve this question


















  • 1





    You're mocking on appSpy but calling doMagic on app...

    – daniu
    Nov 23 '18 at 12:48
















1















I cannot mock a method defined as default in an interface. Can anyone help me here?



The interface has default method providing a logger.



import org.slf4j.Logger;

import org.slf4j.LoggerFactory;



public interface Loggable {



  default Logger logger() {

    return LoggerFactory.getLogger(this.getClass());

  }

}


It is used this way:



public class AppShowOff implements Loggable{



  public void doMagic() {

    logger().debug("It works");

    System.out.println("Works");

  }

}


now I would like to write a test proving that debug method has been called.



import static org.mockito.ArgumentMatchers.any;

import static org.mockito.Mockito.mock;

import static org.mockito.Mockito.times;

import static org.mockito.Mockito.verify;

import static org.mockito.Mockito.when;



import org.junit.Test;

import org.mockito.Mockito;

import org.slf4j.Logger;



public class AppShowOffTest {



  @Test

  public void doMagic() {

    Logger loggerMock = mock(Logger.class);

    Loggable loggableMock = mock(Loggable.class); // <- not needed, but I also tried this way



    // mocks done



    AppShowOff app = new AppShowOff();

    AppShowOff appSpy = Mockito.spy(new AppShowOff());



    when(loggableMock.logger()).thenReturn(loggerMock);

    when(appSpy.logger()).thenReturn(loggerMock);



    app.doMagic();

    verify(loggerMock, times(1)).debug(any());

  }

}


as you can see I have tried to mock the default method in two ways: 



when(loggableMock.logger()).thenReturn(loggerMock);

when(appSpy.logger()).thenReturn(loggerMock);


but it does not work. The result is:




Wanted but not invoked: logger.debug();
-> at so.AppShowOffTest.doMagic(AppShowOffTest.java:29) Actually, there were zero interactions with this mock.







java unit-testing mockito







share|improve this question


















  • 1





    You're mocking on appSpy but calling doMagic on app...

    – daniu
    Nov 23 '18 at 12:48














1












1








1








I cannot mock a method defined as default in an interface. Can anyone help me here?



The interface has default method providing a logger.



import org.slf4j.Logger;

import org.slf4j.LoggerFactory;



public interface Loggable {



  default Logger logger() {

    return LoggerFactory.getLogger(this.getClass());

  }

}


It is used this way:



public class AppShowOff implements Loggable{



  public void doMagic() {

    logger().debug("It works");

    System.out.println("Works");

  }

}


now I would like to write a test proving that debug method has been called.



import static org.mockito.ArgumentMatchers.any;

import static org.mockito.Mockito.mock;

import static org.mockito.Mockito.times;

import static org.mockito.Mockito.verify;

import static org.mockito.Mockito.when;



import org.junit.Test;

import org.mockito.Mockito;

import org.slf4j.Logger;



public class AppShowOffTest {



  @Test

  public void doMagic() {

    Logger loggerMock = mock(Logger.class);

    Loggable loggableMock = mock(Loggable.class); // <- not needed, but I also tried this way



    // mocks done



    AppShowOff app = new AppShowOff();

    AppShowOff appSpy = Mockito.spy(new AppShowOff());



    when(loggableMock.logger()).thenReturn(loggerMock);

    when(appSpy.logger()).thenReturn(loggerMock);



    app.doMagic();

    verify(loggerMock, times(1)).debug(any());

  }

}


as you can see I have tried to mock the default method in two ways: 



when(loggableMock.logger()).thenReturn(loggerMock);

when(appSpy.logger()).thenReturn(loggerMock);


but it does not work. The result is:




Wanted but not invoked: logger.debug();
-> at so.AppShowOffTest.doMagic(AppShowOffTest.java:29) Actually, there were zero interactions with this mock.







java unit-testing mockito







share|improve this question














I cannot mock a method defined as default in an interface. Can anyone help me here?



The interface has default method providing a logger.



import org.slf4j.Logger;

import org.slf4j.LoggerFactory;



public interface Loggable {



  default Logger logger() {

    return LoggerFactory.getLogger(this.getClass());

  }

}


It is used this way:



public class AppShowOff implements Loggable{



  public void doMagic() {

    logger().debug("It works");

    System.out.println("Works");

  }

}


now I would like to write a test proving that debug method has been called.



import static org.mockito.ArgumentMatchers.any;

import static org.mockito.Mockito.mock;

import static org.mockito.Mockito.times;

import static org.mockito.Mockito.verify;

import static org.mockito.Mockito.when;



import org.junit.Test;

import org.mockito.Mockito;

import org.slf4j.Logger;



public class AppShowOffTest {



  @Test

  public void doMagic() {

    Logger loggerMock = mock(Logger.class);

    Loggable loggableMock = mock(Loggable.class); // <- not needed, but I also tried this way



    // mocks done



    AppShowOff app = new AppShowOff();

    AppShowOff appSpy = Mockito.spy(new AppShowOff());



    when(loggableMock.logger()).thenReturn(loggerMock);

    when(appSpy.logger()).thenReturn(loggerMock);



    app.doMagic();

    verify(loggerMock, times(1)).debug(any());

  }

}


as you can see I have tried to mock the default method in two ways: 



when(loggableMock.logger()).thenReturn(loggerMock);

when(appSpy.logger()).thenReturn(loggerMock);


but it does not work. The result is:




Wanted but not invoked: logger.debug();
-> at so.AppShowOffTest.doMagic(AppShowOffTest.java:29) Actually, there were zero interactions with this mock.







java unit-testing mockito




java unit-testing mockito






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 12:45









Panicking DeveloperPanicking Developer

235




235








  • 1





    You're mocking on appSpy but calling doMagic on app...

    – daniu
    Nov 23 '18 at 12:48














  • 1





    You're mocking on appSpy but calling doMagic on app...

    – daniu
    Nov 23 '18 at 12:48








1




1





You're mocking on appSpy but calling doMagic on app...

– daniu
Nov 23 '18 at 12:48





You're mocking on appSpy but calling doMagic on app...

– daniu
Nov 23 '18 at 12:48












1 Answer
1






active

oldest

votes


















0














Here:



AppShowOff app = new AppShowOff();

AppShowOff appSpy = Mockito.spy(new AppShowOff());


That first app ... is never used, besides to call your method under test doMagic().



The simple answer: drop app completely, and invoke appSpy.doMagic().






share|improve this answer
























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    Here:



    AppShowOff app = new AppShowOff();
    
AppShowOff appSpy = Mockito.spy(new AppShowOff());


    That first app ... is never used, besides to call your method under test doMagic().



    The simple answer: drop app completely, and invoke appSpy.doMagic().






    share|improve this answer




























      0














      Here:



      AppShowOff app = new AppShowOff();
      
AppShowOff appSpy = Mockito.spy(new AppShowOff());


      That first app ... is never used, besides to call your method under test doMagic().



      The simple answer: drop app completely, and invoke appSpy.doMagic().






      share|improve this answer


























        0












        0








        0







        Here:



        AppShowOff app = new AppShowOff();
        
AppShowOff appSpy = Mockito.spy(new AppShowOff());


        That first app ... is never used, besides to call your method under test doMagic().



        The simple answer: drop app completely, and invoke appSpy.doMagic().






        share|improve this answer













        Here:



        AppShowOff app = new AppShowOff();
        
AppShowOff appSpy = Mockito.spy(new AppShowOff());


        That first app ... is never used, besides to call your method under test doMagic().



        The simple answer: drop app completely, and invoke appSpy.doMagic().







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 23 '18 at 13:15









        GhostCatGhostCat

        95.5k1794160




        95.5k1794160
































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