Difference between revisions of "Hazardous Objects"

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* 59 - 1 = 58. Get the entry number 58 of the array. It contains 250. C = 250.
 
* 59 - 1 = 58. Get the entry number 58 of the array. It contains 250. C = 250.
 
* 59 > 31, so 59 - 31 = 28. Entry number 28's value is 250. C = 250.
 
* 59 > 31, so 59 - 31 = 28. Entry number 28's value is 250. C = 250.
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==== Invalid byte values ====
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 +
These byte values are converted to 1 (8 mines) by the game:
 +
 +
3, 4, 6, 7, 8, 9, 10, 11, 248, 249, 250, 251, 252, 253, 254, 255.

Revision as of 14:44, 8 October 2011

The Hazardous Objects are the Mines and the Oil Drums (or Barrels).

Object Types

There can be four values : no objects, mines only, oil drums only, or both objects.

Object Count

This value only exists in W:A. You can set how many objects can appear, from 1 to 250. For having zero objects, simply set "no objects" as the type.

Not all values from 1 to 250 are represented. The following is a breakdown of possible values:
1-30 (1 by 1), 35-100 (5 by 5), 110-250 (10 by 10).

How is this setting saved in a WSC file?

Old schemes

In WWP and W:A v1 schemes (and probably W:A v2 old schemes), only the hazardous object types value is saved.

The values are: 0x00=No objects, 0x01=Mines, 0x02=Oil Drums, 0x05=Both.

New schemes

In new W:A schemes, there is the setting Object Count. So both Object Types and Count values are stored in the same byte.

Object Types

You have to modulo (%) the value by 4.

  • If you get 0, there are no objects.
  • If you get 1, there are only mines.
  • If you get 2, there are only oil drums.
  • If you get 3, there are both objects.

For example :

You have the value "79".

79 % 4 = 3 (because 79 / 4 = 19 and the remainder is 3)

So there will be both mines and oil drums in the scheme.

Another example

You have the value "246".

254 % 4 = 2 (because 246 / 4 = 63 and the remainder is 2)

Object Count

First, you will have to re-use the modulo's result as M.

Then, considering V as the byte's value, do :

N = V - (8 + M)

Then finally, divide N by 4 for getting the array's entry (E).

E = N / 4.

Adjust the E value with your array. Meaning for example:

  • You can make an array with a NULL entry as the entry number 0, and you keep the same values for the entries.
  • You can make an array normally, meaning you will have to remove 1 to E.
  • You can make an array with values from 35 to 250 only. If E < 31, E already equals to the value (if E = 5 for example, then C = 5 - considering C is the object count). If E >= 31, then remove 31 to get your array entry and select this entry (for example, 32 - 31 = 1, and 1 is the second value of the array, 1 => 40 so C = 40).

The array's values are, as we said above:

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250.

For example

We have again the value "79". V = 79.

M = 3 as we saw above.

So:

N = 79 - (8 + 3)
N = 79 - 11
N = 68.

Then let's calculate E:

E = 68 / 4
E = 17.

Then let's see with the 3 methods we suggested for the array:

  • Get the 17th entry of the array. Value of it is 17. C = 17.
  • 17 - 1 = 16. Get the entry number 16 of the array. It contains 17. C = 17.
  • 17 < 31, so simply C = E = 17.

Another example

V = 246.

As we said above, M = 2.

So:

N = 246 - (8 + 2)
N = 246 - 10
N = 236.

Then let's calculate E:

E = 236 / 4
E = 59.

Then let's see with the 3 methods we suggested for the array:

  • Get the 59th entry of the array. Value of it is 250. C = 250.
  • 59 - 1 = 58. Get the entry number 58 of the array. It contains 250. C = 250.
  • 59 > 31, so 59 - 31 = 28. Entry number 28's value is 250. C = 250.

Invalid byte values

These byte values are converted to 1 (8 mines) by the game:

3, 4, 6, 7, 8, 9, 10, 11, 248, 249, 250, 251, 252, 253, 254, 255.

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