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.











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.











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.











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






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asked Nov 23 '18 at 12:45









Panicking DeveloperPanicking Developer

82




82








  • 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






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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().






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    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














      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

        90.4k1688146




        90.4k1688146






























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