Pointer to an Integer in Java

Question

I have a code like this

int a,b;
switch(whatever){
    case 1:
      lots_of_lines_dealing_with_variable_a;
    case 2:
      same_lines_but_dealing_with_variable_b;
}

I thought of doing:

int a,b;
pointer_to_int p;
switch(whatever){
    case 1:
      p=a;
    case 2:
      p=b;
}
lots_of_lines_dealing_with_pointer_p;

It would reduce the code to about half the lines, but Java doesn't allow pointers to integers. So, is there any way to approach this?

Edit: The homework is much bigger than just this method. I need to Create a class called "DoubleList" which contains two linked lists in a single Vector. The integers I talk about are the pointers to the start of the Lists, which I need to move to other positions of the Lists when adding or removing elements to the Lists

Answer 1

Maybe I'm missing something, but looks easily solvable to me:

int a,b;

switch(whatever) {
  case 1:
    a = manipulateValue(a);
    break;
  case 2:
    b = manipulateValue(b);
    break;
}

int manipulateValue(int v) {
  // lots of lines dealing with variable v
  return v;
}

If you don't need to modify the variables, then you can leave out the return-value (just use void) and the assignment.

If nothing else needs to call the method, then it should be private (it's a general principle: give as little access as possible, as much as necessary).

Answer 2

You could try using boxing.

public class IntBoxer {
    public IntBoxer() {
    }
    public IntBoxer(int startValue) {
        this.value = startValue;
    }
    public int value;
}
IntBoxer a = new IntBoxer();
IntBoxer b = new IntBoxer();
IntBoxer p;
Switch(whatever){
    case 1:
      p=a;
      break;
    case 2:
      p=b;
      break;
}
lots_of_lines_dealing_with_pointer_p.value;

Answer 3

Yes, it's called "a method":

private int lots_of_lines_dealing_with_value(int x)
{
  .
  .
  .
  return new_value;
}


int a,b;
Switch(whatever){
    case 1:
      p=lots_of_lines_dealing_with_value(a);
      break;
    case 2:
      p=lots_of_lines_dealing_with_value(b);
      break;
}

Answer 4

Use the Integer class instead of an int.

To give you an example using the code you have:

Integer a,b;
Integer p;
switch(whatever){
    case 1:
      p=a;
    case 2:
      p=b;
}

Answer 5

Why can't you just do:

int a,b;

Switch(whatever){
    case 1:
      f(a);
      break;
    case 2:
      f(b);
      break;
    default:
      // ???
}

Answer 6

Java doesn't have pointers, it has references.

A pointer is a variable that contains a memory address. Typically one dereferences the pointer to obtain the memory needed for some operation.

A reference is an index into a memory management table. Typically the memory management table is protected from direct access. In Java's case the reference can't be manipulated manually, as any read of the variable will result in returning the referenced object.

This has a lot of implications, but is necessary for decent automatic garbage collection. Garbage collection sometimes involves moving objects in memory to create larger areas of free space (for needed objects that cannot fit into the current memory holes). If Java exposed a pointer, then after memory compaction, you might have the "old" address of the memory.

By using references, your reference is guaranteed to stay the same, even if the actual location of memory moves around. Internal to the JVM is a reference to pointer table, but you will never get to see it from a running program; because, if you ever touched it, it would mess up the automatic memory management.

Answer 7

Well, for this scenario:

boolean useA;
int a, b;

if(useA)
      lots_of_lines_dealing_with_variable_a;
else
      same_lines_but_dealing_with_variable_b;

you can use:

boolean useA;
int a, b;

int value = useA ? a : b;

// lines_dealing_with_value

if(useA)
    a = value;
else
    b = value;

But really. Methods!