/* * 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. */ #include PUBLIC __allrem /* FUNCTIONS ***************************************************************/ .code // // 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 END