Dry Underflow
Posted by: luckytyphlosion
Date: 2015-01-27 16:38:18
This is basically a method of item underflow, except you don't need the guard to take the fresh water from your inventory. (hence the name "dry underflow", since there isn't a freshwater involved. Credits for Stumpdotio for coming up with this name)
EDIT: At this point, I realize that you can underflow items using the above method.
[size=14pt]New Method:[/size]
Requirements:
- [li]2 tossable items without 255 quantity[/li]
[li]1 tossable item with 255 quantity[/li]
[li]NO KEY ITEMS IN INVENTORY[/li]
Procedure
[li]Order items so that your 2 tossable items are in slots 1 and 2 and 255 item is in slot 3[/li]
[li]Toss everything below the 255 item[/li]
[li]Toss the top two items so you have 3 255 items[/li]
Item Setup
Item Count: 1
Item3 x255
Item3 x255
Item3 x255
Cancel x0 (quantity of Cancel doesn't matter)
After acquiring the above setup, which should look like what is shown above, toss 253 of the first 255 stack.
Toss 253
Item Count: 1
Item3 x2
Item3 x255
Item3 x255
Cancel x0
Then, swap the x2 Stack with the x255 Stack.
Swap
Item Count: 0
Item3 x1
Item3 x255
Cancel x0
Swap again to achieve Item Underflow.
Swap again
Item Count: 255
Item3 x0
Cancel x0
[size=14pt]Old Method:[/size]
Requirements:
- [li]2 tossable items without 255 quantity[/li]
[li]1 tossable item with 255 quantity[/li]
[li]NO KEY ITEMS IN INVENTORY[/li]
Method:
[li]Order items so that your 2 tossable items are in slots 1 and 2 and 255 item is in slot 3[/li]
[li]Toss everything below the 255 item[/li]
[li]Toss the top two items so you have 3 255 items[/li]
[li]Swap items 1 and 2 and slot 2 will become 99 (x5, 99, 255)[/li]
[li]Toss all but one of item 1 (1, 99, 255)[/li]
[li]Swap item 2 and 1 (99, 1, 255) (YOU MUST SWAP 2 with 1 NOT 1 with 2, THIS IS THE ONLY PART YOU MIGHT SCREW UP)[/li]
[li]Toss all but one of item 1 again (1, 1, 255)[/li]
[li]Swap item 1 and 2 (2, 255)[/li]
[li]Swap item 1 and 2 again[/li]
[size=14pt]Explanation of the Old Method[/size]
While this isn't related to the New Method, this does explains the mechanics of the glitch that are used in the new method.
When you deposit the three items into the PC (the two useless items and the 255 item stack), and you toss the two items, the item count decrements (read: is subtracted by one). However, the 255 stack shifts up instead of the Cancel Button, so by tossing the two items, you get this layout in your box items.
(note: the item names are simply labels, and do not represent the position of items in the box)
Before Tossing
Box Items Quantity: 3
Item1 x1 (quantity doesn't matter)
Item2 x1 (same case as above)
Item3 x255
Cancel x0 (quantity doesn't matter)
After Tossing Item1 and Item2 in your box items
Box Items Quantity: 1
Item3 x255
Item3 x255
Item3 x255
Cancel x0
This means that you can swap whatever is in the Cancel Slot with the item above. Right now, it doesn't seem useful.
With the current setup, you can't simply toss the remaining 255 stacks, since the Cancel Button Item doesn't actually back you out of the menu. Instead, the Number of Items + 1 determines the location of the real Cancel Button.
Right now, our goal is to decrement the amount of items in the box when the number of items in the box is 0, so that it will underflow to 255. You can't simply toss items, as stated above. To do this, we need to create a setup to decrement the item counter when our item count is 0.
If you swap the two stacks of 255 you can currently access, then you end up with:
Item3 x156
Item3 x99
Item3 x255
Then, we toss 155 of the first item so we have one remaining.
Item3 x1
Item3 x99
Item3 x255
Now, the next part here is very weird. If you swap Item3 x1 with Item3 x99, nothing happens. However, if you swap Item3 x99 with Item3 x1, then Item3 x99 is swapped with Item3 x1.
Item3 x99
Item3 x1
Item3 x255
We can now toss x98 of the first stack, so that only one remains. Then, we can swap both x1 Stacks with each other, decrementing the item count so we have 0 items.
Item Count: 0
Item3 x2
Item3 x255
Normally, you wouldn't be able to scroll down since you have 0 items. For some reason though, if you decrement the item counter BY swapping an item, the cursor boundaries will not change. This allows us to swap Item3 x2 with Item3 x255, decrementing the item counter once again, and allowing you to access other bytes in memory.
Item Count: 255
Item3 x1 (since 255 + 2 = 1, because of overflow)