pocketquake e-100 not working

[wildag]

I am new to this but here goes. I have installed everything (as far as I can tell) correctly (but obviously I havn't). I have a 64mb storage card so here is how I have it all laid out.<br><br>\Storage Card\quake\ID1\pak0.pak.gz <br>\quake\pocketquake.exe<br>\windows\gx.dll<br>\quake\ID1\autoexec.cfg<br><br>I get an error message saying 'PocketQuake (or one of its components can not be found). What does that mean/what do I do? Sorry if this is a simple question, but I am really stumped.

[Dan East]

Sorry, only Pocket PCs and HPC/Pro devices are supported at this time. <br><br>Dan East

[wildag]

I have a Casio Cassiopeia e-100... that's not a pocket pc? Has the MIPS processor... same thing as the E-105 and so forth....<br><br>I was under the impression that that is the same thing.

[suchiaruzu]

And an E-105 isnt a PocketPC either. PocketPCs run Windows CE 3.0, the ones you mentioned run CE 2.11

[Dan East]

Those devices are officially known as "Palm-Sized PCs".<br><br>Dan East

[wildag]

Oh, I see.. so its the windows CE 2.11 that's not supported. Not the fact that its a E-100 machine. Right? (Or am I confused about this?)<br><br>Please forgive me... I'm used to the Palm world...<br><br>BTW... if it is not supported... I would love to provide 'support' for it. I am a C and Java programer. Just tell me what I need (libraries, dev programs, etc) and I will try to come up with a solution. I think it would help me learn alot about these things. Any suggestions you have as far as 'development' sites or 'tutorials' for programing would be great. Thank-you very much.

[Moose or Chuck]

A stdio.lib needs to be created for the WinCE 2.11 OS, and the game api needs to be ported.<br><br>To clear up the confusion:<br>There are many different versions of the WinCE OS:<br>WinCE 2.11 devices = PalmPCs (E-105, E-100, etc.)<br>WinCE 3.0 devices = PocketPCs (iPaq, E-125, Jornada 525, etc.)

[SkaBoss]

Yes someone else, a Jeremy Higgenbothom (if thats how you spell it) was going to then dissappeared (thats another story) so if you are going to try you have my support.  I just wish I could program well enough to help.

[wildag]

well, no promisses, but I can certainly try

[Matt Keys]

feel free to post your questions about it in the developer discussion, many people would be more than happy to help you out, and they almost always have a solution.

[R0B]

Of course we always have a solution.  Although deviding by 0 doesn't always help, it is a solution.

[Moose or Chuck]

You know, I'm not too sure dividing by zero is infinity anymore. It could be null (meaning, undefinable; as opposed to 0).

[Diego Cueva]

You know, other day I said to my teacher that I could divide a number by zero... She said "no", but I was to bored to give her an explanation, so I just stayed quiet.<br><br><br><br><br><br><br><br>That was todays bored post.<br>Thank you,<br>Diego Cueva

[Moose or Chuck]

You know, we never did settle that div by 0 problem...<br><br>Hrmm<br><br>*starts to think about it*<br>*hmmmm, pies. Infinite pies!*<br>*err... head _ hurts!*<br><br>

[R0B]

Moose, in a way you are right, but in a way you are not.  here is the proof.<br>5 / 0 = x Given<br>5 / 0 * 0 = x * 0 Multiplicitive prop. of =<br>5 = x * 0 Property of Multiplication (I think)<br>5 = 0 Same as above.<br><br>This proves that it is an emty set, but look at this.<br><br>(5 / 0) = x Given<br>0 *(5 / 0)= x * 0 Multiplicitive prop. of =<br>0 = x * 0 distrubutive prop.<br><br>0 = 0 Property of multiplication<br><br>So, with that case, dividing a num by 0 gives you zero, here is why it is infinite.<br><br>10 / 0 = 0~0 Prop of division<br><br>0 goes into 10 0~0 times.  This simply shows that the # zero goes into 10 infinite times.  The reason why it is infinite and not empty set is because of the property of division that states that when one number is divided by another number the result is the amount of times that the second number can be subtracted from the first number untill it reaches 0 (I think that it is worded close to that).  So, it really depends on how you simplify an equation as to how you interprite 0<br><br>Now here is another question.<br>1/9 = .1111111111~<br>2/9 = .2222222222~<br>and so on<br>That is why some people think that 9/9 = .99999999~, and that .99999999~ = 1.  But look at this.<br><br><br>.99999999~ + (.000000000~1  * 10 ) = 1<br>(note that the one is after the vinculum (is that how that is spelled?)in the second number .000000000~1)<br><br>That would mean that .000000000~1  * 10 has no value in math if that equ is wrong, or if it is right, then .9999999~ just happens to be really close to one, but not quite there.  Tell me what you think.