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how to call outp (conio.h)

Posted on 2001-07-12
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Last Modified: 2008-02-01
I get a
LNK2001 unresolved external symbol _outp
link error on the following code:

#include "conio.h"

int main (void)
{
  outp(0x64, 0xFE); //reboot Win95
  return 0;
}

Can somebody help me to get this code running?
I am trying to reproduce assembler
OUT 0x64, 0xFE
code to perform a Win95 reboot.
0
Comment
Question by:saracoglu
7 Comments
 
LVL 86

Expert Comment

by:jkr
ID: 6278716
Why don't you use

ExitWindowsEx ( EXW_REBOOT, NULL);

instead?
0
 
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Expert Comment

by:jhance
ID: 6278759
Basically, you cannot call the outp instruction from a WIN32 user-mode application.  It's a protected instruction and can only be executed from the privileged instruction level.

jkr has the correct suggestion to reboot Windows from a WIN32 app.  If you really need to do outp, you need a VxD or static device driver or you need to "thunk" to 16-bit mode.
0
 
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Author Comment

by:saracoglu
ID: 6278760
I need to reboot win95 on an almost crashed situation. The ExitWindowsEx API does not work then.
Somebody has compiled me a VC dll to call from VB. It works perfect.
He sent me only this code piece, and said, that was it.
Now he is on holiday, and I need to reproduce this code.
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LVL 7

Expert Comment

by:KangaRoo
ID: 6278860
You may have to compile and link the code for 16 bit DOS mode (target = Dos executable, /not/ win32 console)
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Author Comment

by:saracoglu
ID: 6278894
I have found it.
I needed to put an underscore before outp (like: _outp)

thanks for the ideas.
The code is now working as follows:

#define DllExport __declspec( dllexport )

#include <conio.h>

extern "C" DllExport long Reboot9x()
{
 int Dummy;
 Dummy = _outp(0x64, 0xFE);
 return 0;
}

I will contact ee to move this question to paq
0
 
LVL 3

Expert Comment

by:JackThornton
ID: 6279293
jhance:

Actually, you can do ins and outs in Win32, at least from Win95. It is expensive, because when the priviledged instruction is trapped by the operating system, Win95 (at least) reproduces the in or out in ring0 before returning to the ring3 code. Moral is you can do it, but it is ***slow***.

- jack
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Accepted Solution

by:
modder earned 0 total points
ID: 6292832
As you found your own solution, I'll refund your points, and I'll make this a PAQ

modder
Community Support Moderator@Experts-Exchange
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