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[Release] Steed Walking

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Old 05/03/2011, 19:39   #16
 
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Quote:
Originally Posted by _DreadNought_ View Post
like
xi = x2
yi = y2

Cant make it more simple then that, its just naming.

Good job spirited, Big difference to Angulius's(?) release.

Thanks mate, glad to see you back. As you know if you ever need help hit me up on msn.
Well, I'm leaving the switch statement up because the other method was not my idea and the switch statement shows how the directions relate to walking better. It's not the simplest way to do it though and there are probably many other ways to do it- but the array idea is pretty **** cool, and I think I'm going to use it on many other things besides walking that I've been using switch statements or unnecessary dictionaries *in dictionaries* for.

*edited



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Old 05/04/2011, 21:29   #17
 
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Then comment out the switch part in your code and use the array, so you get both readability and efficiency. Though IMO the array method is easier to read and understand.


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Old 05/05/2011, 00:52   #18
 
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Originally Posted by KraHen View Post
Then comment out the switch part in your code and use the array, so you get both readability and efficiency. Though IMO the array method is easier to read and understand.
I'm not editing my post. I didn't come up with it, so I shouldn't take the credit. I put a link below it instead.
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Old 05/05/2011, 14:13   #19
 
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to save some ppl time use this...

Code:
        public static sbyte[] NormalX = new sbyte[8] { 0, -1, -1, -1, 0, 1, 1, 1 };
        public static sbyte[] NormalY = new sbyte[8] { 1, 1, 0, -1, -1, -1, 0, 1 };
        public static sbyte[] SteedX = new sbyte[24] { 0, -2, -2, -2, 0, 2, 2, 2, -1, -2, -2, -1, 1, 2, 2, 1, -1, -2, -2, -1, 1, 2, 2, 1 };
        public static sbyte[] SteedY = new sbyte[24] { 2, 2, 0, -2, -2, -2, 0, 2, 2, 1, -1, -2, -2, -1, 1, 2, 2, 1, -1, -2, -2, -1, 1, 2 };
cya!


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