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Problem 2 - Code comprehension

    public class Foo {
        public int x;
        public int y;

        public Foo(int a) {
            x = a;
            y = 1;
        }

        public int bar(int a) {
            x += a;
            y += 1;
            return (x + y);
        }

        public void bar(Foo f) {
            if (f.x < x) {
                f.bar(x);
            } else {
                bar(f.x);
            }
        }
    }

Fill in the table based on the following

1: Foo f1 = new Foo(1);
2: Foo f2 = new Foo(2);
3: f1.bar(3);
4: f2.bar(4);
5: f1.bar(f2.bar(5));
6: f2.bar(f1);
7: f1.bar(new Foo(6));

So we are going to walk through the values after every call.

By the time we get to line 3 we have the following information

Line 3 gets executed: f1.bar(3);

• we are editing the values for the f1 varible
• bar(3) is called. so x gets increased by 3 and y by 1
• we don’t care about the return statement because nothing happens.

Line 4: f2.bar(4);

• we are editing the f2 variable now
• bar(4) is called. so x gets increased by 4 and y by 1
• we ignore the return statement again

line 5: f1.bar(f2.bar(5));

• Lets start by doing the inside: f2.bar(5)
  • editing the f2 variable
  • x increases by 5 and y by 1
  • Now f2: x = 11 and y = 3
  • And now we care about the return statement because it is used in the nested call.
  • return (x + y) == 14
• now we are calling f1.bar(14)
• editing the f1 variable
• x increases by 14 and y by 1
• Now f1: x = 18 and y = 3
• don’t care about the return statement this time

Line 6: f2.bar(f1);

• now we have to look at what happens when we call bar with a Foo object.
• its saying if f1.x < f2.x we then call f1.bar(f2.x);
• else just call f2.bar(f1.x);
• It is important to remember which class you are operating in. It makes it easier to rewrite the code in terms of f1 and f2 so you can make the right calls with the right values.
• In this case we are going to go with the else block
  • so we execute f2.bar(f1.x);
  • we modify f2 and increase its x by f1.x and increase y by 1
  • x = 11 + 18 = 29 and y = 4
  • ignore return statement

line 7: f1.bar(new Foo(6));

• our new Foo has an x = 6 and y = 1;
• just like in our previous block we have:
  if newfoo.x < f1.x we then call newfoo.bar(f1.x);
  else just call f1.bar(newfoo.x);
• in this case we would execute the if block. The if block modifies the object newfoo that we made but this object has no impact on us and is discarded after the expression is returned. So it doesn’t change f1 or f2