EasyMock:如何在没有警告的情况下创建泛化类的模拟？

   <dependency>
       <groupId>org.easymock</groupId>
       <artifactId>easymock</artifactId>
       <version>4.4</version>
       <scope>test</scope>
    </dependency>
   <dependency>
       <groupId>junit</groupId>
       <artifactId>junit</artifactId>
       <version>4.13.2</version>
       <scope>test</scope>
    </dependency>
</dependencies>

public interface MyGenericInterface<T> {
    T getData();
}

import org.easymock.EasyMock;
import org.junit.Before;
import org.junit.Test;

import static org.junit.Assert.assertEquals;

public class MyGenericInterfaceTest {
    @Mock
    private MyGenericInterface<String> myGenericInterface;

    @Before
    public void setUp() {
        myGenericInterface = EasyMock.createMock(MyGenericInterface.class);
    }

    @Test
    public void testGetData() {
        EasyMock.expect(myGenericInterface.getData()).andReturn("Hello, world!");
        EasyMock.replay(myGenericInterface);

        String result = myGenericInterface.getData();
        assertEquals("Hello, world!", result);

        EasyMock.verify(myGenericInterface);
    }
}