code by Michael Avila,
written by Kirupa  3 April 2006
In the previous
page, I briefly framed the problem associated with
getting our checkCombination method to work. Let's continue
our stepbystep trace of how it works in this page.
First, let's go through the nested for loop:
 for
(var
j:Number
= 0;
j<key_combinations.length;
j++)
{
 for
(var
i:Number
=
0; i<keys_pressed.length;
i++)
{

if
(keys_pressed[i]
==
key_combinations[j][i+1])
{

if
(i
==
key_combinations[j].length2)
{

invokeOnKeyCombination(key_combinations[j][0]);

return;
 }
 }
else
{

break;
 }
 }
 }
Initially, j is initialized to zero. In the second loop,
i is initialized to 0 also. We are now at the if statement:
if (keys_pressed[0] ==
key_combinations[0][1])
I substituted a 0 for both the i and j variables in the
if statement. keys_pressed[0] maps to the number 17 from our
keys_pressed array. If you recall, The key_combinations
array is a nested array. By calling key_combinations[0][1],
I am calling the first array's second item  which is 17.
Success!
Because the previous if statement equated out to true, we
go to the second if statement (with i and j values
substituted in):
if (0 ==
key_combinations[0].length  2)
The length of our key_combinations[0] array ([redo, 17,
89]), is three. So, 32 is 1, and 0 does not equal 1. This
conditional results in false. Let's go back to our second
for loop and try again!
Trial 0.1
So, we are back at the second for loop. The value for j
remains the same at 0, because the first for loop is what
controls the value of the j variable. But the value of i
will increment (i++). It is now 1. Now, our first if
statement is:
if (keys_pressed[1] ==
key_combinations[0][2])
The maps out to if (90 == 89). Nope  that does not work!
We jump down the break statement and exit out of this loop.
We are now back at the first for loop. Let's try again!
Trial 1.0
So, we are back at (almost) where we started from  the
first for loop. This time, the value of j is set to 1
instead instead of 0. We proceed to the second for loop, and
we start back with i being equal to 1. Now, let's take a
look at the first if statement again:
if (keys_pressed[0] ==
key_combinations[1][1])
Notice that the value of j is 1, so we will be looking at
the second item in our key_combinations array. More
specifically, we will be looking at key_combinations' second
item's second item:
key_combinations' second item:
[undo, 17, 90]
key_combinations' second item's second item:
17
Similar to before, our if statement checks out.
keys_pressed[0] is 17 and so is the value returned by our
key_combinations[1][1] array. We then go to the second if
statement:
if (0 ==
key_combinations[1].length  2)
Like a recurring bad dream, though, you will
unfortunately have the equality 0 == 1, which is false. Even
though our j value is incremented by one, it still refers to
an item of length 3.
Phew! We are still not done yet. Let's cover some more of
our checkCombinations code on the
next page
and wrap this tutorial up!
