doubt

[meena latha]

public class call
{
public void m1()
{
System.out.println("matt");
}
}
 
public class B extends call
{
public static void main(String[]arhg)
{call a = new B();
 a.m1();
}
public void m1()
{
System.out.println("mbtt");
}
}

if there is a question like this and the answer include
1.compile error
2.mbtt
3.matt
which one you will select.

[ February 28, 2005: Message edited by: ramya jp ]
[ March 01, 2005: Message edited by: ramya jp ]

[ramaseshan T]

"mbtt" as it is holding the subclass object.

[reshmi Bijo]

Hi Ramya

The ans is 2)mbtt because the method invocation depends on the type of the object. Here the object is of class B.

[meena latha]

why not compile error since it is having two public class....???

[ankur rathi]

Waw , Good point ...

But it is not mentioned here that , this two classes are in same source file .

[meena latha]

hi rathi...so if it not mentioned that they are in same source code file then whether we have to eliminate the compile error option???
Thanks in advance.

[Jeroen Wenting]

On the real exam you will know whether two pieces of code are in the same source by the line numbering.
This will also tell you whether any source you see is a complete class or a fragment.