RE: from where comes "__moddi3"?

2007-07-21 Thread Arjan van de Ven
On Sat, 2007-07-21 at 14:27 -0700, David Schwartz wrote:
> > On 7/21/07, Jan Engelhardt <[EMAIL PROTECTED]> wrote:
> > > v & 0x0F  <=>  v % 16
> > Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.
> 
> Actually, it's the compiler's bad. That's a pretty fundamental equivalence 
> that the compiler should recognize for native integral types.

for unsigned ones for sure yes... but the original question was about 15
not 16.


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RE: from where comes "__moddi3"?

2007-07-21 Thread David Schwartz

> On 7/21/07, Jan Engelhardt <[EMAIL PROTECTED]> wrote:
> > v & 0x0F  <=>  v % 16
> Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.

Actually, it's the compiler's bad. That's a pretty fundamental equivalence that 
the compiler should recognize for native integral types.

DS


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Re: from where comes "__moddi3"?

2007-07-21 Thread Peter Zijlstra
On Sat, 2007-07-21 at 21:58 +0200, Jan-Benedict Glaw wrote:
> On Sat, 2007-07-21 21:31:13 +0200, Peter Zijlstra <[EMAIL PROTECTED]> wrote:
> > On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
> > 
> > use do_div().
> > 
> > 16 works because gcc translates that into a right shift.
> 
> Right shift?  Doesn't it just mask out everything but the low 4 bit?

Yeah, its bitwise and for modulo and shifts for mult and div.

Silly mistake on my side.

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Re: from where comes "__moddi3"?

2007-07-21 Thread Alexander Shishkin

On 7/21/07, Jan Engelhardt <[EMAIL PROTECTED]> wrote:

v & 0x0F  <=>  v % 16

Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.

Regards,
--
Alex
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Re: from where comes "__moddi3"?

2007-07-21 Thread Jan-Benedict Glaw
On Sat, 2007-07-21 21:31:13 +0200, Peter Zijlstra <[EMAIL PROTECTED]> wrote:
> On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
> 
> use do_div().
> 
> 16 works because gcc translates that into a right shift.

Right shift?  Doesn't it just mask out everything but the low 4 bit?

MfG, JBG

-- 
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Re: from where comes "__moddi3"?

2007-07-21 Thread Robert P. J. Day
On Sat, 21 Jul 2007, Peter Zijlstra wrote:

> On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
> > again, probably displaying my abject ignorance, but i wrote a
> > trivial module that tries to "var % 15", and i get:
> >
> >   WARNING: "__moddi3" undefined!
> >
> > and, not surprisingly, when i try to insmod:
> >
> >   insmod: error inserting 'seq.ko': -1 Unknown symbol in module
> >
> > (using 16 rather than 15 works fine, as i assume that the modulus
> > call is simply replaced by an optimized  bitwise comparison.)
> >
> >   so ... from where comes __moddi3?  i know there are places in the
> > kernel source tree that do non-power-of-2 moduli, and they work fine,
> > no?  thanks.
>
> let me guess, 32 bit kernel, and var is 64 bit?

um ... yeah.  duh.  var type is actually loff_t.  i should have
realized that.

> gcc translates that into a libgcc call, which we _explicitly_ do not
> have because 64bit divisions are expensive!
>
> use do_div().

gotcha.  thanks.

rday
-- 

Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

http://fsdev.net/wiki/index.php?title=Main_Page

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Re: from where comes "__moddi3"?

2007-07-21 Thread Jan Engelhardt

On Jul 21 2007 23:33, Alexander Shishkin wrote:
> On 7/21/07, Robert P. J. Day <[EMAIL PROTECTED]> wrote:
>>
>>   again, probably displaying my abject ignorance, but i wrote a
>> trivial module that tries to "var % 15", and i get:
>>
>>   WARNING: "__moddi3" undefined!
> ...which comes from libgcc1 which you obviously don't want to link against.
>
> Does (var & 0x0f) not work for you?

v & 0x0F  <=>  v % 16



Jan
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Re: from where comes "__moddi3"?

2007-07-21 Thread Alexander Shishkin

On 7/21/07, Robert P. J. Day <[EMAIL PROTECTED]> wrote:


  again, probably displaying my abject ignorance, but i wrote a
trivial module that tries to "var % 15", and i get:

  WARNING: "__moddi3" undefined!

...which comes from libgcc1 which you obviously don't want to link against.

Does (var & 0x0f) not work for you?

Regards,
--
Alex
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Re: from where comes "__moddi3"?

2007-07-21 Thread Peter Zijlstra
On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
> again, probably displaying my abject ignorance, but i wrote a
> trivial module that tries to "var % 15", and i get:
> 
>   WARNING: "__moddi3" undefined!
> 
> and, not surprisingly, when i try to insmod:
> 
>   insmod: error inserting 'seq.ko': -1 Unknown symbol in module
> 
> (using 16 rather than 15 works fine, as i assume that the modulus
> call is simply replaced by an optimized  bitwise comparison.)
> 
>   so ... from where comes __moddi3?  i know there are places in the
> kernel source tree that do non-power-of-2 moduli, and they work fine,
> no?  thanks.

let me guess, 32 bit kernel, and var is 64 bit?

gcc translates that into a libgcc call, which we _explicitly_ do not
have because 64bit divisions are expensive!

use do_div().

16 works because gcc translates that into a right shift.

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from where comes "__moddi3"?

2007-07-21 Thread Robert P. J. Day

  again, probably displaying my abject ignorance, but i wrote a
trivial module that tries to "var % 15", and i get:

  WARNING: "__moddi3" undefined!

and, not surprisingly, when i try to insmod:

  insmod: error inserting 'seq.ko': -1 Unknown symbol in module

(using 16 rather than 15 works fine, as i assume that the modulus
call is simply replaced by an optimized  bitwise comparison.)

  so ... from where comes __moddi3?  i know there are places in the
kernel source tree that do non-power-of-2 moduli, and they work fine,
no?  thanks.

rday
-- 

Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

http://fsdev.net/wiki/index.php?title=Main_Page

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from where comes __moddi3?

2007-07-21 Thread Robert P. J. Day

  again, probably displaying my abject ignorance, but i wrote a
trivial module that tries to var % 15, and i get:

  WARNING: __moddi3 undefined!

and, not surprisingly, when i try to insmod:

  insmod: error inserting 'seq.ko': -1 Unknown symbol in module

(using 16 rather than 15 works fine, as i assume that the modulus
call is simply replaced by an optimized  bitwise comparison.)

  so ... from where comes __moddi3?  i know there are places in the
kernel source tree that do non-power-of-2 moduli, and they work fine,
no?  thanks.

rday
-- 

Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

http://fsdev.net/wiki/index.php?title=Main_Page

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Re: from where comes __moddi3?

2007-07-21 Thread Peter Zijlstra
On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
 again, probably displaying my abject ignorance, but i wrote a
 trivial module that tries to var % 15, and i get:
 
   WARNING: __moddi3 undefined!
 
 and, not surprisingly, when i try to insmod:
 
   insmod: error inserting 'seq.ko': -1 Unknown symbol in module
 
 (using 16 rather than 15 works fine, as i assume that the modulus
 call is simply replaced by an optimized  bitwise comparison.)
 
   so ... from where comes __moddi3?  i know there are places in the
 kernel source tree that do non-power-of-2 moduli, and they work fine,
 no?  thanks.

let me guess, 32 bit kernel, and var is 64 bit?

gcc translates that into a libgcc call, which we _explicitly_ do not
have because 64bit divisions are expensive!

use do_div().

16 works because gcc translates that into a right shift.

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Re: from where comes __moddi3?

2007-07-21 Thread Alexander Shishkin

On 7/21/07, Robert P. J. Day [EMAIL PROTECTED] wrote:


  again, probably displaying my abject ignorance, but i wrote a
trivial module that tries to var % 15, and i get:

  WARNING: __moddi3 undefined!

...which comes from libgcc1 which you obviously don't want to link against.

Does (var  0x0f) not work for you?

Regards,
--
Alex
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Re: from where comes __moddi3?

2007-07-21 Thread Jan Engelhardt

On Jul 21 2007 23:33, Alexander Shishkin wrote:
 On 7/21/07, Robert P. J. Day [EMAIL PROTECTED] wrote:

   again, probably displaying my abject ignorance, but i wrote a
 trivial module that tries to var % 15, and i get:

   WARNING: __moddi3 undefined!
 ...which comes from libgcc1 which you obviously don't want to link against.

 Does (var  0x0f) not work for you?

v  0x0F  =  v % 16



Jan
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Re: from where comes __moddi3?

2007-07-21 Thread Robert P. J. Day
On Sat, 21 Jul 2007, Peter Zijlstra wrote:

 On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
  again, probably displaying my abject ignorance, but i wrote a
  trivial module that tries to var % 15, and i get:
 
WARNING: __moddi3 undefined!
 
  and, not surprisingly, when i try to insmod:
 
insmod: error inserting 'seq.ko': -1 Unknown symbol in module
 
  (using 16 rather than 15 works fine, as i assume that the modulus
  call is simply replaced by an optimized  bitwise comparison.)
 
so ... from where comes __moddi3?  i know there are places in the
  kernel source tree that do non-power-of-2 moduli, and they work fine,
  no?  thanks.

 let me guess, 32 bit kernel, and var is 64 bit?

um ... yeah.  duh.  var type is actually loff_t.  i should have
realized that.

 gcc translates that into a libgcc call, which we _explicitly_ do not
 have because 64bit divisions are expensive!

 use do_div().

gotcha.  thanks.

rday
-- 

Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

http://fsdev.net/wiki/index.php?title=Main_Page

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Re: from where comes __moddi3?

2007-07-21 Thread Jan-Benedict Glaw
On Sat, 2007-07-21 21:31:13 +0200, Peter Zijlstra [EMAIL PROTECTED] wrote:
 On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
 
 use do_div().
 
 16 works because gcc translates that into a right shift.

Right shift?  Doesn't it just mask out everything but the low 4 bit?

MfG, JBG

-- 
  Jan-Benedict Glaw  [EMAIL PROTECTED]  +49-172-7608481
Signature of: God put me on earth to accomplish a certain number of
the second  :things. Right now I am so far behind I will never die.


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Re: from where comes __moddi3?

2007-07-21 Thread Alexander Shishkin

On 7/21/07, Jan Engelhardt [EMAIL PROTECTED] wrote:

v  0x0F  =  v % 16

Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.

Regards,
--
Alex
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Re: from where comes __moddi3?

2007-07-21 Thread Peter Zijlstra
On Sat, 2007-07-21 at 21:58 +0200, Jan-Benedict Glaw wrote:
 On Sat, 2007-07-21 21:31:13 +0200, Peter Zijlstra [EMAIL PROTECTED] wrote:
  On Sat, 2007-07-21 at 15:21 -0400, Robert P. J. Day wrote:
  
  use do_div().
  
  16 works because gcc translates that into a right shift.
 
 Right shift?  Doesn't it just mask out everything but the low 4 bit?

Yeah, its bitwise and for modulo and shifts for mult and div.

Silly mistake on my side.

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RE: from where comes __moddi3?

2007-07-21 Thread David Schwartz

 On 7/21/07, Jan Engelhardt [EMAIL PROTECTED] wrote:
  v  0x0F  =  v % 16
 Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.

Actually, it's the compiler's bad. That's a pretty fundamental equivalence that 
the compiler should recognize for native integral types.

DS


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RE: from where comes __moddi3?

2007-07-21 Thread Arjan van de Ven
On Sat, 2007-07-21 at 14:27 -0700, David Schwartz wrote:
  On 7/21/07, Jan Engelhardt [EMAIL PROTECTED] wrote:
   v  0x0F  =  v % 16
  Indeed. (Why would anyone want to mod/div by 15 anyway?). My bad.
 
 Actually, it's the compiler's bad. That's a pretty fundamental equivalence 
 that the compiler should recognize for native integral types.

for unsigned ones for sure yes... but the original question was about 15
not 16.


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