Build it step by step

Every step below leaves a scenario that compiles and runs. If you stop halfway you will have something that works, not something broken. Follow along in Greenfoot rather than reading straight through.

1. Step 1

   On your screen The standard traversal loop with the index used inside the body.

   int[] lanes = {3, 7, 2, 9};
   for (int i = 0; i < lanes.length; i++)
   {
       addObject(new Coin(), lanes[i], 0);
   }

   This is the shape. Commit it to memory as one thing:

   `for (int i = 0; i < a.length; i++)`

   Start at 0 because that is the first index. Stop BEFORE length because the last index is length minus one. Step by one to visit each in turn.

   Inside, `lanes[i]` is the element at this pass.

2. Step 2

   On your screen A traversal that hardcodes the size and misses an element after the array is changed.

   int[] lanes = {3, 7, 2, 9, 5};   // now five
   for (int i = 0; i < 4; i++)      // still says four
   {
       // the 5 is never visited
   }

   Why `lanes.length` rather than 4? Because arrays change during development.

   Add a fifth value and this loop silently misses it. No error, just an element that is never processed. Write `<= 4` instead and you get the opposite failure, an exception.

   `i < a.length` is correct at every size, forever, and needs no maintenance.

3. Step 3

   On your screen A loop filling an array with computed values.

   int[] squares = new int[5];
   for (int i = 0; i < squares.length; i++)
   {
       squares[i] = i * i;
   }
   // {0, 1, 4, 9, 16}

   Filling rather than reading. Same loop, with the assignment the other way round.

   And here is the counter-in-a-calculation idea from 4.7 again: `i * i` turns the index into a value. Unit 6 uses exactly this to turn an index into a screen position.