a problem with the resolt of the code

[alex lotel]

this is how i think this code should work
i put it also in the debugger
and it also shows the resolt 11 2 7

and then when it goes to the print method suddenly it shows
a=7 b=2 c=7

why??

i showed here the diagram which represents my logic

http://img401.imageshack.us/my.php?image=img8298gp7.jpg

where did i go wrong??

public class A implements C
{
int d=17;
A(int d)
{
this.d=d;
}
A done()
{
System.out.println("I am done");
return this;
}
}
public class B extends A 
{
int a=1;
int b=2;
public B(int a, int b)
{
this(a);
a=b;
}
public B(int b)
{
super(b);
a=b;
}
public A done()
{
System.out.println("finished");
return this;
}
 
public void print(int a, int b, int c)
{
System.out.println("a=" +a);
System.out.println("b=" +b);
System.out.println("c=" +c);
}
}
public interface C 
{
public void print(int a, int b, int c);
}
 
public class MainClass
 {
 
public static void main(String[] args) 
{
B b=new B(7,11);
b.print(b.a,b.b,b.d);
A a= ((A)b).done();

}
 
}

[Campbell Ritchie]

I hope you never hand in such convoluted code for an examination.

You are setting the parameter a equal to 11 in the two-arguments constructor, not the field, which would be this.a

[Henry Wong]

First of all, this code should not compile. Class A doesn't implement interface C. But, assuming that you posted the wrong code, and everything else is correct...

this is how i think this code should work
i put it also in the debugger
and it also shows the resolt 11 2 7

Well, I highly doubt that... There are no timing issues (shown in the code) related to this application that would make an actual run different from a debugging run.

and then when it goes to the print method suddenly it shows
a=7 b=2 c=7

why??

Well, 7 2 7 is correct. And here is why...

public B(int a, int b)
{
    this(a);
    a=b;                      // This line (Step 6) in your diagram
}

In step 6 of your diagram, you have this line as assigning the value of the parameter b to the instance variable a. This is not true. This line assigns the value of the parameter b to the parameter variable a. To do what you want expect in step 6, you need to do this...

public B(int a, int b)
{
    this(a);
    this.a=b;
}

Henry

[alex lotel]

i forgot to mention that the first part of the question was
to fix the code

i added abstract in the the signature of class A

is that changes some how your answers??

[Henry Wong]

Originally posted by donaldth smithts:

is that changes some how your answers??

That explains why it compiles. None of the other answers changes.

Henry

[alex lotel]

so you are saying that in step 6

that i changed the input variable "a"(a variable which exists only in the signature)??
and not this.a??

my next question (presumption) is that in step 5 i made a similar
command
but here b.a do change although its a=b too
because the signature is public B(int b)
there is no a there so we change the b.a variable
am i correct
[ February 17, 2008: Message edited by: donaldth smithts ]

[Campbell Ritchie]

Originally posted by donaldth smithts:
so you are saying . . .
my next question (presumption) is that in step 5 i made a similar
command
but here b.a do change although its a=b too
because the signature is public B(int b)
there is no a there so we change the b.a variable
am i correct?

Yes.