How to mock an interface default method Java8/Mockito2

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.

asked Nov 23 '18 at 12:45

Panicking Developer

235

1

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

– daniu
Nov 23 '18 at 12:48

add a comment |

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.

asked Nov 23 '18 at 12:45

Panicking Developer

235

1

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

– daniu
Nov 23 '18 at 12:48

add a comment |

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.

asked Nov 23 '18 at 12:45

Panicking Developer

235

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

asked Nov 23 '18 at 12:45

Panicking Developer

235

asked Nov 23 '18 at 12:45

Panicking Developer

235

asked Nov 23 '18 at 12:45

Panicking Developer

235

asked Nov 23 '18 at 12:45

Panicking Developer

235

asked Nov 23 '18 at 12:45

Panicking Developer

235

1

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

– daniu
Nov 23 '18 at 12:48

add a comment |

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

add a comment |

1 Answer
1

active

oldest

votes

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().

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

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().

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

add a comment |

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().

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

add a comment |

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().

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

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().

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

answered Nov 23 '18 at 13:15

GhostCat

95.5k1794160

add a comment |

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