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Timo Kreuzer 2010-02-26 22:57:55 +00:00
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lib/sdk/crt/math/i386/allrem_asm.s Normal file
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/*
* COPYRIGHT: See COPYING in the top level directory
* PROJECT: ReactOS kernel
* PURPOSE: Run-Time Library
* FILE: lib/rtl/i386/allrem.S
* PROGRAMER: Alex Ionescu (alex@relsoft.net)
*
* Copyright (C) 2002 Michael Ringgaard.
* All rights reserved.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
*
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
* 3. Neither the name of the project nor the names of its contributors
* may be used to endorse or promote products derived from this software
* without specific prior written permission.
* THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS "AS IS" AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT OWNER OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES// LOSS OF USE, DATA, OR PROFITS// OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*/
.globl __allrem
.intel_syntax noprefix
/* FUNCTIONS ***************************************************************/
//
// llrem - signed long remainder
//
// Purpose:
// Does a signed long remainder of the arguments. Arguments are
// not changed.
//
// Entry:
// Arguments are passed on the stack:
// 1st pushed: divisor (QWORD)
// 2nd pushed: dividend (QWORD)
//
// Exit:
// EDX:EAX contains the remainder (dividend%divisor)
// NOTE: this routine removes the parameters from the stack.
//
// Uses:
// ECX
//
__allrem :
push ebx
push edi
// Set up the local stack and save the index registers. When this is done
// the stack frame will look as follows (assuming that the expression a%b will
// generate a call to lrem(a, b)):
//
// -----------------
// | |
// |---------------|
// | |
// |--divisor (b)--|
// | |
// |---------------|
// | |
// |--dividend (a)-|
// | |
// |---------------|
// | return addr** |
// |---------------|
// | EBX |
// |---------------|
// ESP---->| EDI |
// -----------------
//
#undef DVNDLO
#undef DVNDHI
#undef DVSRLO
#undef DVSRHI
#define DVNDLO [esp + 12] // stack address of dividend (a)
#define DVNDHI [esp + 16] // stack address of dividend (a)
#define DVSRLO [esp + 20] // stack address of divisor (b)
#define DVSRHI [esp + 24] // stack address of divisor (b)
// Determine sign of the result (edi = 0 if result is positive, non-zero
// otherwise) and make operands positive.
xor edi,edi // result sign assumed positive
mov eax,DVNDHI // hi word of a
or eax,eax // test to see if signed
jge short .L1 // skip rest if a is already positive
inc edi // complement result sign flag bit
mov edx,DVNDLO // lo word of a
neg eax // make a positive
neg edx
sbb eax,0
mov DVNDHI,eax // save positive value
mov DVNDLO,edx
.L1:
mov eax,DVSRHI // hi word of b
or eax,eax // test to see if signed
jge short .L2 // skip rest if b is already positive
mov edx,DVSRLO // lo word of b
neg eax // make b positive
neg edx
sbb eax,0
mov DVSRHI,eax // save positive value
mov DVSRLO,edx
.L2:
//
// Now do the divide. First look to see if the divisor is less than 4194304K.
// If so, then we can use a simple algorithm with word divides, otherwise
// things get a little more complex.
//
// NOTE - eax currently contains the high order word of DVSR
//
or eax,eax // check to see if divisor < 4194304K
jnz short .L3 // nope, gotta do this the hard way
mov ecx,DVSRLO // load divisor
mov eax,DVNDHI // load high word of dividend
xor edx,edx
div ecx // edx <- remainder
mov eax,DVNDLO // edx:eax <- remainder:lo word of dividend
div ecx // edx <- final remainder
mov eax,edx // edx:eax <- remainder
xor edx,edx
dec edi // check result sign flag
jns short .L4 // negate result, restore stack and return
jmp short .L8 // result sign ok, restore stack and return
//
// Here we do it the hard way. Remember, eax contains the high word of DVSR
//
.L3:
mov ebx,eax // ebx:ecx <- divisor
mov ecx,DVSRLO
mov edx,DVNDHI // edx:eax <- dividend
mov eax,DVNDLO
.L5:
shr ebx,1 // shift divisor right one bit
rcr ecx,1
shr edx,1 // shift dividend right one bit
rcr eax,1
or ebx,ebx
jnz short .L5 // loop until divisor < 4194304K
div ecx // now divide, ignore remainder
//
// We may be off by one, so to check, we will multiply the quotient
// by the divisor and check the result against the orignal dividend
// Note that we must also check for overflow, which can occur if the
// dividend is close to 2**64 and the quotient is off by 1.
//
mov ecx,eax // save a copy of quotient in ECX
mul dword ptr DVSRHI
xchg ecx,eax // save product, get quotient in EAX
mul dword ptr DVSRLO
add edx,ecx // EDX:EAX = QUOT * DVSR
jc short .L6 // carry means Quotient is off by 1
//
// do long compare here between original dividend and the result of the
// multiply in edx:eax. If original is larger or equal, we are ok, otherwise
// subtract the original divisor from the result.
//
cmp edx,DVNDHI // compare hi words of result and original
ja short .L6 // if result > original, do subtract
jb short .L7 // if result < original, we are ok
cmp eax,DVNDLO // hi words are equal, compare lo words
jbe short .L7 // if less or equal we are ok, else subtract
.L6:
sub eax,DVSRLO // subtract divisor from result
sbb edx,DVSRHI
.L7:
//
// Calculate remainder by subtracting the result from the original dividend.
// Since the result is already in a register, we will do the subtract in the
// opposite direction and negate the result if necessary.
//
sub eax,DVNDLO // subtract dividend from result
sbb edx,DVNDHI
//
// Now check the result sign flag to see if the result is supposed to be positive
// or negative. It is currently negated (because we subtracted in the 'wrong'
// direction), so if the sign flag is set we are done, otherwise we must negate
// the result to make it positive again.
//
dec edi // check result sign flag
jns short .L8 // result is ok, restore stack and return
.L4:
neg edx // otherwise, negate the result
neg eax
sbb edx,0
//
// Just the cleanup left to do. edx:eax contains the quotient.
// Restore the saved registers and return.
//
.L8:
pop edi
pop ebx
ret 16