948 lines
30 KiB
C++
948 lines
30 KiB
C++
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#ifdef _M_IX86 // use this file only for 32-bit architecture
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#define CRT_LOWORD(x) dword ptr [x+0]
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#define CRT_HIWORD(x) dword ptr [x+4]
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extern "C"
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{
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__declspec(naked) void _alldiv()
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{
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#define DVND esp + 16 // stack address of dividend (a)
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#define DVSR esp + 24 // stack address of divisor (b)
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__asm
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{
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push edi
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push esi
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push ebx
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; Determine sign of the result (edi = 0 if result is positive, non-zero
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; otherwise) and make operands positive.
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xor edi,edi ; result sign assumed positive
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mov eax,CRT_HIWORD(DVND) ; hi word of a
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or eax,eax ; test to see if signed
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jge short L1 ; skip rest if a is already positive
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inc edi ; complement result sign flag
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mov edx,CRT_LOWORD(DVND) ; lo word of a
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neg eax ; make a positive
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neg edx
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sbb eax,0
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mov CRT_HIWORD(DVND),eax ; save positive value
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mov CRT_LOWORD(DVND),edx
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L1:
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mov eax,CRT_HIWORD(DVSR) ; hi word of b
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or eax,eax ; test to see if signed
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jge short L2 ; skip rest if b is already positive
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inc edi ; complement the result sign flag
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mov edx,CRT_LOWORD(DVSR) ; lo word of a
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neg eax ; make b positive
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neg edx
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sbb eax,0
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mov CRT_HIWORD(DVSR),eax ; save positive value
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mov CRT_LOWORD(DVSR),edx
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L2:
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;
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; Now do the divide. First look to see if the divisor is less than 4194304K.
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; If so, then we can use a simple algorithm with word divides, otherwise
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; things get a little more complex.
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;
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; NOTE - eax currently contains the high order word of DVSR
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;
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or eax,eax ; check to see if divisor < 4194304K
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jnz short L3 ; nope, gotta do this the hard way
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mov ecx,CRT_LOWORD(DVSR) ; load divisor
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mov eax,CRT_HIWORD(DVND) ; load high word of dividend
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xor edx,edx
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div ecx ; eax <- high order bits of quotient
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mov ebx,eax ; save high bits of quotient
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mov eax,CRT_LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
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div ecx ; eax <- low order bits of quotient
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mov edx,ebx ; edx:eax <- quotient
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jmp short L4 ; set sign, restore stack and return
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;
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; Here we do it the hard way. Remember, eax contains the high word of DVSR
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;
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L3:
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mov ebx,eax ; ebx:ecx <- divisor
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mov ecx,CRT_LOWORD(DVSR)
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mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
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mov eax,CRT_LOWORD(DVND)
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L5:
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shr ebx,1 ; shift divisor right one bit
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rcr ecx,1
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shr edx,1 ; shift dividend right one bit
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rcr eax,1
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or ebx,ebx
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jnz short L5 ; loop until divisor < 4194304K
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div ecx ; now divide, ignore remainder
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mov esi,eax ; save quotient
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;
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; We may be off by one, so to check, we will multiply the quotient
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; by the divisor and check the result against the orignal dividend
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; Note that we must also check for overflow, which can occur if the
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; dividend is close to 2**64 and the quotient is off by 1.
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;
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mul CRT_HIWORD(DVSR) ; QUOT * CRT_HIWORD(DVSR)
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mov ecx,eax
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mov eax,CRT_LOWORD(DVSR)
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mul esi ; QUOT * CRT_LOWORD(DVSR)
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add edx,ecx ; EDX:EAX = QUOT * DVSR
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jc short L6 ; carry means Quotient is off by 1
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;
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; do long compare here between original dividend and the result of the
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; multiply in edx:eax. If original is larger or equal, we are ok, otherwise
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; subtract one (1) from the quotient.
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;
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cmp edx,CRT_HIWORD(DVND) ; compare hi words of result and original
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ja short L6 ; if result > original, do subtract
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jb short L7 ; if result < original, we are ok
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cmp eax,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
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jbe short L7 ; if less or equal we are ok, else subtract
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L6:
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dec esi ; subtract 1 from quotient
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L7:
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xor edx,edx ; edx:eax <- quotient
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mov eax,esi
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;
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; Just the cleanup left to do. edx:eax contains the quotient. Set the sign
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; according to the save value, cleanup the stack, and return.
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;
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L4:
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dec edi ; check to see if result is negative
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jnz short L8 ; if EDI == 0, result should be negative
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neg edx ; otherwise, negate the result
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neg eax
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sbb edx,0
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;
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; Restore the saved registers and return.
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;
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L8:
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pop ebx
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pop esi
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pop edi
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ret 16
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}
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#undef DVND
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#undef DVSR
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}
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__declspec(naked) void _alldvrm()
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{
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#define DVND esp + 16 // stack address of dividend (a)
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#define DVSR esp + 24 // stack address of divisor (b)
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__asm
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{
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push edi
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push esi
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push ebp
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; Determine sign of the quotient (edi = 0 if result is positive, non-zero
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; otherwise) and make operands positive.
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; Sign of the remainder is kept in ebp.
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xor edi,edi ; result sign assumed positive
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xor ebp,ebp ; result sign assumed positive
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mov eax,CRT_HIWORD(DVND) ; hi word of a
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or eax,eax ; test to see if signed
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jge short L1 ; skip rest if a is already positive
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inc edi ; complement result sign flag
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inc ebp ; complement result sign flag
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mov edx,CRT_LOWORD(DVND) ; lo word of a
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neg eax ; make a positive
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neg edx
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sbb eax,0
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mov CRT_HIWORD(DVND),eax ; save positive value
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mov CRT_LOWORD(DVND),edx
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L1:
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mov eax,CRT_HIWORD(DVSR) ; hi word of b
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or eax,eax ; test to see if signed
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jge short L2 ; skip rest if b is already positive
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inc edi ; complement the result sign flag
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mov edx,CRT_LOWORD(DVSR) ; lo word of a
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neg eax ; make b positive
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neg edx
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sbb eax,0
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mov CRT_HIWORD(DVSR),eax ; save positive value
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mov CRT_LOWORD(DVSR),edx
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L2:
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;
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; Now do the divide. First look to see if the divisor is less than 4194304K.
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||
|
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; If so, then we can use a simple algorithm with word divides, otherwise
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||
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; things get a little more complex.
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||
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;
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; NOTE - eax currently contains the high order word of DVSR
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;
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or eax,eax ; check to see if divisor < 4194304K
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jnz short L3 ; nope, gotta do this the hard way
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mov ecx,CRT_LOWORD(DVSR) ; load divisor
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mov eax,CRT_HIWORD(DVND) ; load high word of dividend
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xor edx,edx
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div ecx ; eax <- high order bits of quotient
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mov ebx,eax ; save high bits of quotient
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mov eax,CRT_LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
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div ecx ; eax <- low order bits of quotient
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mov esi,eax ; ebx:esi <- quotient
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;
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; Now we need to do a multiply so that we can compute the remainder.
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;
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mov eax,ebx ; set up high word of quotient
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mul CRT_LOWORD(DVSR) ; CRT_HIWORD(QUOT) * DVSR
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mov ecx,eax ; save the result in ecx
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mov eax,esi ; set up low word of quotient
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mul CRT_LOWORD(DVSR) ; CRT_LOWORD(QUOT) * DVSR
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add edx,ecx ; EDX:EAX = QUOT * DVSR
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jmp short L4 ; complete remainder calculation
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;
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; Here we do it the hard way. Remember, eax contains the high word of DVSR
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;
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L3:
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mov ebx,eax ; ebx:ecx <- divisor
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mov ecx,CRT_LOWORD(DVSR)
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mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
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mov eax,CRT_LOWORD(DVND)
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L5:
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shr ebx,1 ; shift divisor right one bit
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rcr ecx,1
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shr edx,1 ; shift dividend right one bit
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rcr eax,1
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or ebx,ebx
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jnz short L5 ; loop until divisor < 4194304K
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div ecx ; now divide, ignore remainder
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mov esi,eax ; save quotient
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;
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; We may be off by one, so to check, we will multiply the quotient
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||
|
|
; by the divisor and check the result against the orignal dividend
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||
|
|
; Note that we must also check for overflow, which can occur if the
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||
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; dividend is close to 2**64 and the quotient is off by 1.
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;
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mul CRT_HIWORD(DVSR) ; QUOT * CRT_HIWORD(DVSR)
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mov ecx,eax
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mov eax,CRT_LOWORD(DVSR)
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mul esi ; QUOT * CRT_LOWORD(DVSR)
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add edx,ecx ; EDX:EAX = QUOT * DVSR
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jc short L6 ; carry means Quotient is off by 1
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;
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; do long compare here between original dividend and the result of the
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; multiply in edx:eax. If original is larger or equal, we are ok, otherwise
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; subtract one (1) from the quotient.
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;
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cmp edx,CRT_HIWORD(DVND) ; compare hi words of result and original
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ja short L6 ; if result > original, do subtract
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jb short L7 ; if result < original, we are ok
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cmp eax,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
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jbe short L7 ; if less or equal we are ok, else subtract
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L6:
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dec esi ; subtract 1 from quotient
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sub eax,CRT_LOWORD(DVSR) ; subtract divisor from result
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sbb edx,CRT_HIWORD(DVSR)
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L7:
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xor ebx,ebx ; ebx:esi <- quotient
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L4:
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;
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; Calculate remainder by subtracting the result from the original dividend.
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; Since the result is already in a register, we will do the subtract in the
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; opposite direction and negate the result if necessary.
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;
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sub eax,CRT_LOWORD(DVND) ; subtract dividend from result
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sbb edx,CRT_HIWORD(DVND)
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;
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; Now check the result sign flag to see if the result is supposed to be positive
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; or negative. It is currently negated (because we subtracted in the 'wrong'
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; direction), so if the sign flag is set we are done, otherwise we must negate
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; the result to make it positive again.
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;
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dec ebp ; check result sign flag
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jns short L9 ; result is ok, set up the quotient
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neg edx ; otherwise, negate the result
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neg eax
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sbb edx,0
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;
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; Now we need to get the quotient into edx:eax and the remainder into ebx:ecx.
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;
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L9:
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mov ecx,edx
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mov edx,ebx
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mov ebx,ecx
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mov ecx,eax
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mov eax,esi
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;
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||
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; Just the cleanup left to do. edx:eax contains the quotient. Set the sign
|
||
|
|
; according to the save value, cleanup the stack, and return.
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||
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;
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||
|
|
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dec edi ; check to see if result is negative
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jnz short L8 ; if EDI == 0, result should be negative
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neg edx ; otherwise, negate the result
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neg eax
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sbb edx,0
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;
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; Restore the saved registers and return.
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;
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L8:
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pop ebp
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pop esi
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pop edi
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ret 16
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}
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#undef DVND
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#undef DVSR
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}
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||
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__declspec(naked) void _allmul()
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{
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#define A esp + 8 // stack address of a
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#define B esp + 16 // stack address of b
|
||
|
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||
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__asm
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||
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|
{
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push ebx
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mov eax,CRT_HIWORD(A)
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mov ecx,CRT_LOWORD(B)
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mul ecx ;eax has AHI, ecx has BLO, so AHI * BLO
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mov ebx,eax ;save result
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||
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mov eax,CRT_LOWORD(A)
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mul CRT_HIWORD(B) ;ALO * BHI
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add ebx,eax ;ebx = ((ALO * BHI) + (AHI * BLO))
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mov eax,CRT_LOWORD(A) ;ecx = BLO
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mul ecx ;so edx:eax = ALO*BLO
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add edx,ebx ;now edx has all the LO*HI stuff
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||
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pop ebx
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ret 16 ; callee restores the stack
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||
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}
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||
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#undef A
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#undef B
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}
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||
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||
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__declspec(naked) void _allrem()
|
||
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|
{
|
||
|
|
#define DVND esp + 12 // stack address of dividend (a)
|
||
|
|
#define DVSR esp + 20 // stack address of divisor (b)
|
||
|
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|
||
|
|
__asm
|
||
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|
{
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||
|
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push ebx
|
||
|
|
push edi
|
||
|
|
|
||
|
|
|
||
|
|
; 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,CRT_HIWORD(DVND) ; 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,CRT_LOWORD(DVND) ; lo word of a
|
||
|
|
neg eax ; make a positive
|
||
|
|
neg edx
|
||
|
|
sbb eax,0
|
||
|
|
mov CRT_HIWORD(DVND),eax ; save positive value
|
||
|
|
mov CRT_LOWORD(DVND),edx
|
||
|
|
L1:
|
||
|
|
mov eax,CRT_HIWORD(DVSR) ; hi word of b
|
||
|
|
or eax,eax ; test to see if signed
|
||
|
|
jge short L2 ; skip rest if b is already positive
|
||
|
|
mov edx,CRT_LOWORD(DVSR) ; lo word of b
|
||
|
|
neg eax ; make b positive
|
||
|
|
neg edx
|
||
|
|
sbb eax,0
|
||
|
|
mov CRT_HIWORD(DVSR),eax ; save positive value
|
||
|
|
mov CRT_LOWORD(DVSR),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,CRT_LOWORD(DVSR) ; load divisor
|
||
|
|
mov eax,CRT_HIWORD(DVND) ; load high word of dividend
|
||
|
|
xor edx,edx
|
||
|
|
div ecx ; edx <- remainder
|
||
|
|
mov eax,CRT_LOWORD(DVND) ; 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,CRT_LOWORD(DVSR)
|
||
|
|
mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
|
||
|
|
mov eax,CRT_LOWORD(DVND)
|
||
|
|
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 CRT_HIWORD(DVSR)
|
||
|
|
xchg ecx,eax ; save product, get quotient in EAX
|
||
|
|
mul CRT_LOWORD(DVSR)
|
||
|
|
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,CRT_HIWORD(DVND) ; 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,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
|
||
|
|
jbe short L7 ; if less or equal we are ok, else subtract
|
||
|
|
L6:
|
||
|
|
sub eax,CRT_LOWORD(DVSR) ; subtract divisor from result
|
||
|
|
sbb edx,CRT_HIWORD(DVSR)
|
||
|
|
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,CRT_LOWORD(DVND) ; subtract dividend from result
|
||
|
|
sbb edx,CRT_HIWORD(DVND)
|
||
|
|
|
||
|
|
;
|
||
|
|
; 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
|
||
|
|
}
|
||
|
|
|
||
|
|
#undef DVND
|
||
|
|
#undef DVSR
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _allshl()
|
||
|
|
{
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
;
|
||
|
|
; Handle shifts of 64 or more bits (all get 0)
|
||
|
|
;
|
||
|
|
cmp cl, 64
|
||
|
|
jae short RETZERO
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 0 and 31 bits
|
||
|
|
;
|
||
|
|
cmp cl, 32
|
||
|
|
jae short MORE32
|
||
|
|
shld edx,eax,cl
|
||
|
|
shl eax,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 32 and 63 bits
|
||
|
|
;
|
||
|
|
MORE32:
|
||
|
|
mov edx,eax
|
||
|
|
xor eax,eax
|
||
|
|
and cl,31
|
||
|
|
shl edx,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; return 0 in edx:eax
|
||
|
|
;
|
||
|
|
RETZERO:
|
||
|
|
xor eax,eax
|
||
|
|
xor edx,edx
|
||
|
|
ret
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _allshr()
|
||
|
|
{
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
;
|
||
|
|
; Handle shifts of 64 bits or more (if shifting 64 bits or more, the result
|
||
|
|
; depends only on the high order bit of edx).
|
||
|
|
;
|
||
|
|
cmp cl,64
|
||
|
|
jae short RETSIGN
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 0 and 31 bits
|
||
|
|
;
|
||
|
|
cmp cl, 32
|
||
|
|
jae short MORE32
|
||
|
|
shrd eax,edx,cl
|
||
|
|
sar edx,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 32 and 63 bits
|
||
|
|
;
|
||
|
|
MORE32:
|
||
|
|
mov eax,edx
|
||
|
|
sar edx,31
|
||
|
|
and cl,31
|
||
|
|
sar eax,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; Return double precision 0 or -1, depending on the sign of edx
|
||
|
|
;
|
||
|
|
RETSIGN:
|
||
|
|
sar edx,31
|
||
|
|
mov eax,edx
|
||
|
|
ret
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _aulldiv()
|
||
|
|
{
|
||
|
|
#define DVND esp + 12 // stack address of dividend (a)
|
||
|
|
#define DVSR esp + 20 // stack address of divisor (b)
|
||
|
|
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
push ebx
|
||
|
|
push esi
|
||
|
|
|
||
|
|
;
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
mov eax,CRT_HIWORD(DVSR) ; check to see if divisor < 4194304K
|
||
|
|
or eax,eax
|
||
|
|
jnz short L1 ; nope, gotta do this the hard way
|
||
|
|
mov ecx,CRT_LOWORD(DVSR) ; load divisor
|
||
|
|
mov eax,CRT_HIWORD(DVND) ; load high word of dividend
|
||
|
|
xor edx,edx
|
||
|
|
div ecx ; get high order bits of quotient
|
||
|
|
mov ebx,eax ; save high bits of quotient
|
||
|
|
mov eax,CRT_LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
|
||
|
|
div ecx ; get low order bits of quotient
|
||
|
|
mov edx,ebx ; edx:eax <- quotient hi:quotient lo
|
||
|
|
jmp short L2 ; restore stack and return
|
||
|
|
|
||
|
|
;
|
||
|
|
; Here we do it the hard way. Remember, eax contains DVSRHI
|
||
|
|
;
|
||
|
|
|
||
|
|
L1:
|
||
|
|
mov ecx,eax ; ecx:ebx <- divisor
|
||
|
|
mov ebx,CRT_LOWORD(DVSR)
|
||
|
|
mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
|
||
|
|
mov eax,CRT_LOWORD(DVND)
|
||
|
|
L3:
|
||
|
|
shr ecx,1 ; shift divisor right one bit; hi bit <- 0
|
||
|
|
rcr ebx,1
|
||
|
|
shr edx,1 ; shift dividend right one bit; hi bit <- 0
|
||
|
|
rcr eax,1
|
||
|
|
or ecx,ecx
|
||
|
|
jnz short L3 ; loop until divisor < 4194304K
|
||
|
|
div ebx ; now divide, ignore remainder
|
||
|
|
mov esi,eax ; save quotient
|
||
|
|
|
||
|
|
;
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
mul CRT_HIWORD(DVSR) ; QUOT * CRT_HIWORD(DVSR)
|
||
|
|
mov ecx,eax
|
||
|
|
mov eax,CRT_LOWORD(DVSR)
|
||
|
|
mul esi ; QUOT * CRT_LOWORD(DVSR)
|
||
|
|
add edx,ecx ; EDX:EAX = QUOT * DVSR
|
||
|
|
jc short L4 ; 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 one (1) from the quotient.
|
||
|
|
;
|
||
|
|
|
||
|
|
cmp edx,CRT_HIWORD(DVND) ; compare hi words of result and original
|
||
|
|
ja short L4 ; if result > original, do subtract
|
||
|
|
jb short L5 ; if result < original, we are ok
|
||
|
|
cmp eax,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
|
||
|
|
jbe short L5 ; if less or equal we are ok, else subtract
|
||
|
|
L4:
|
||
|
|
dec esi ; subtract 1 from quotient
|
||
|
|
L5:
|
||
|
|
xor edx,edx ; edx:eax <- quotient
|
||
|
|
mov eax,esi
|
||
|
|
|
||
|
|
;
|
||
|
|
; Just the cleanup left to do. edx:eax contains the quotient.
|
||
|
|
; Restore the saved registers and return.
|
||
|
|
;
|
||
|
|
|
||
|
|
L2:
|
||
|
|
|
||
|
|
pop esi
|
||
|
|
pop ebx
|
||
|
|
|
||
|
|
ret 16
|
||
|
|
}
|
||
|
|
|
||
|
|
#undef DVND
|
||
|
|
#undef DVSR
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _aulldvrm()
|
||
|
|
{
|
||
|
|
#define DVND esp + 8 // stack address of dividend (a)
|
||
|
|
#define DVSR esp + 16 // stack address of divisor (b)
|
||
|
|
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
push esi
|
||
|
|
|
||
|
|
;
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
mov eax,CRT_HIWORD(DVSR) ; check to see if divisor < 4194304K
|
||
|
|
or eax,eax
|
||
|
|
jnz short L1 ; nope, gotta do this the hard way
|
||
|
|
mov ecx,CRT_LOWORD(DVSR) ; load divisor
|
||
|
|
mov eax,CRT_HIWORD(DVND) ; load high word of dividend
|
||
|
|
xor edx,edx
|
||
|
|
div ecx ; get high order bits of quotient
|
||
|
|
mov ebx,eax ; save high bits of quotient
|
||
|
|
mov eax,CRT_LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
|
||
|
|
div ecx ; get low order bits of quotient
|
||
|
|
mov esi,eax ; ebx:esi <- quotient
|
||
|
|
|
||
|
|
;
|
||
|
|
; Now we need to do a multiply so that we can compute the remainder.
|
||
|
|
;
|
||
|
|
mov eax,ebx ; set up high word of quotient
|
||
|
|
mul CRT_LOWORD(DVSR) ; CRT_HIWORD(QUOT) * DVSR
|
||
|
|
mov ecx,eax ; save the result in ecx
|
||
|
|
mov eax,esi ; set up low word of quotient
|
||
|
|
mul CRT_LOWORD(DVSR) ; CRT_LOWORD(QUOT) * DVSR
|
||
|
|
add edx,ecx ; EDX:EAX = QUOT * DVSR
|
||
|
|
jmp short L2 ; complete remainder calculation
|
||
|
|
|
||
|
|
;
|
||
|
|
; Here we do it the hard way. Remember, eax contains DVSRHI
|
||
|
|
;
|
||
|
|
|
||
|
|
L1:
|
||
|
|
mov ecx,eax ; ecx:ebx <- divisor
|
||
|
|
mov ebx,CRT_LOWORD(DVSR)
|
||
|
|
mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
|
||
|
|
mov eax,CRT_LOWORD(DVND)
|
||
|
|
L3:
|
||
|
|
shr ecx,1 ; shift divisor right one bit; hi bit <- 0
|
||
|
|
rcr ebx,1
|
||
|
|
shr edx,1 ; shift dividend right one bit; hi bit <- 0
|
||
|
|
rcr eax,1
|
||
|
|
or ecx,ecx
|
||
|
|
jnz short L3 ; loop until divisor < 4194304K
|
||
|
|
div ebx ; now divide, ignore remainder
|
||
|
|
mov esi,eax ; save quotient
|
||
|
|
|
||
|
|
;
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
mul CRT_HIWORD(DVSR) ; QUOT * CRT_HIWORD(DVSR)
|
||
|
|
mov ecx,eax
|
||
|
|
mov eax,CRT_LOWORD(DVSR)
|
||
|
|
mul esi ; QUOT * CRT_LOWORD(DVSR)
|
||
|
|
add edx,ecx ; EDX:EAX = QUOT * DVSR
|
||
|
|
jc short L4 ; 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 one (1) from the quotient.
|
||
|
|
;
|
||
|
|
|
||
|
|
cmp edx,CRT_HIWORD(DVND) ; compare hi words of result and original
|
||
|
|
ja short L4 ; if result > original, do subtract
|
||
|
|
jb short L5 ; if result < original, we are ok
|
||
|
|
cmp eax,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
|
||
|
|
jbe short L5 ; if less or equal we are ok, else subtract
|
||
|
|
L4:
|
||
|
|
dec esi ; subtract 1 from quotient
|
||
|
|
sub eax,CRT_LOWORD(DVSR) ; subtract divisor from result
|
||
|
|
sbb edx,CRT_HIWORD(DVSR)
|
||
|
|
L5:
|
||
|
|
xor ebx,ebx ; ebx:esi <- quotient
|
||
|
|
|
||
|
|
L2:
|
||
|
|
;
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
sub eax,CRT_LOWORD(DVND) ; subtract dividend from result
|
||
|
|
sbb edx,CRT_HIWORD(DVND)
|
||
|
|
neg edx ; otherwise, negate the result
|
||
|
|
neg eax
|
||
|
|
sbb edx,0
|
||
|
|
|
||
|
|
;
|
||
|
|
; Now we need to get the quotient into edx:eax and the remainder into ebx:ecx.
|
||
|
|
;
|
||
|
|
mov ecx,edx
|
||
|
|
mov edx,ebx
|
||
|
|
mov ebx,ecx
|
||
|
|
mov ecx,eax
|
||
|
|
mov eax,esi
|
||
|
|
;
|
||
|
|
; Just the cleanup left to do. edx:eax contains the quotient.
|
||
|
|
; Restore the saved registers and return.
|
||
|
|
;
|
||
|
|
|
||
|
|
pop esi
|
||
|
|
|
||
|
|
ret 16
|
||
|
|
}
|
||
|
|
|
||
|
|
#undef DVND
|
||
|
|
#undef DVSR
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _aullrem()
|
||
|
|
{
|
||
|
|
#define DVND esp + 8 // stack address of dividend (a)
|
||
|
|
#define DVSR esp + 16 // stack address of divisor (b)
|
||
|
|
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
push ebx
|
||
|
|
|
||
|
|
; 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.
|
||
|
|
;
|
||
|
|
|
||
|
|
mov eax,CRT_HIWORD(DVSR) ; check to see if divisor < 4194304K
|
||
|
|
or eax,eax
|
||
|
|
jnz short L1 ; nope, gotta do this the hard way
|
||
|
|
mov ecx,CRT_LOWORD(DVSR) ; load divisor
|
||
|
|
mov eax,CRT_HIWORD(DVND) ; load high word of dividend
|
||
|
|
xor edx,edx
|
||
|
|
div ecx ; edx <- remainder, eax <- quotient
|
||
|
|
mov eax,CRT_LOWORD(DVND) ; edx:eax <- remainder:lo word of dividend
|
||
|
|
div ecx ; edx <- final remainder
|
||
|
|
mov eax,edx ; edx:eax <- remainder
|
||
|
|
xor edx,edx
|
||
|
|
jmp short L2 ; restore stack and return
|
||
|
|
|
||
|
|
;
|
||
|
|
; Here we do it the hard way. Remember, eax contains DVSRHI
|
||
|
|
;
|
||
|
|
|
||
|
|
L1:
|
||
|
|
mov ecx,eax ; ecx:ebx <- divisor
|
||
|
|
mov ebx,CRT_LOWORD(DVSR)
|
||
|
|
mov edx,CRT_HIWORD(DVND) ; edx:eax <- dividend
|
||
|
|
mov eax,CRT_LOWORD(DVND)
|
||
|
|
L3:
|
||
|
|
shr ecx,1 ; shift divisor right one bit; hi bit <- 0
|
||
|
|
rcr ebx,1
|
||
|
|
shr edx,1 ; shift dividend right one bit; hi bit <- 0
|
||
|
|
rcr eax,1
|
||
|
|
or ecx,ecx
|
||
|
|
jnz short L3 ; loop until divisor < 4194304K
|
||
|
|
div ebx ; 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 CRT_HIWORD(DVSR)
|
||
|
|
xchg ecx,eax ; put partial product in ECX, get quotient in EAX
|
||
|
|
mul CRT_LOWORD(DVSR)
|
||
|
|
add edx,ecx ; EDX:EAX = QUOT * DVSR
|
||
|
|
jc short L4 ; 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're ok, otherwise
|
||
|
|
; subtract the original divisor from the result.
|
||
|
|
;
|
||
|
|
|
||
|
|
cmp edx,CRT_HIWORD(DVND) ; compare hi words of result and original
|
||
|
|
ja short L4 ; if result > original, do subtract
|
||
|
|
jb short L5 ; if result < original, we're ok
|
||
|
|
cmp eax,CRT_LOWORD(DVND) ; hi words are equal, compare lo words
|
||
|
|
jbe short L5 ; if less or equal we're ok, else subtract
|
||
|
|
L4:
|
||
|
|
sub eax,CRT_LOWORD(DVSR) ; subtract divisor from result
|
||
|
|
sbb edx,CRT_HIWORD(DVSR)
|
||
|
|
L5:
|
||
|
|
|
||
|
|
;
|
||
|
|
; Calculate remainder by subtracting the result from the original dividend.
|
||
|
|
; Since the result is already in a register, we will perform the subtract in
|
||
|
|
; the opposite direction and negate the result to make it positive.
|
||
|
|
;
|
||
|
|
|
||
|
|
sub eax,CRT_LOWORD(DVND) ; subtract original dividend from result
|
||
|
|
sbb edx,CRT_HIWORD(DVND)
|
||
|
|
neg edx ; and negate it
|
||
|
|
neg eax
|
||
|
|
sbb edx,0
|
||
|
|
|
||
|
|
;
|
||
|
|
; Just the cleanup left to do. dx:ax contains the remainder.
|
||
|
|
; Restore the saved registers and return.
|
||
|
|
;
|
||
|
|
|
||
|
|
L2:
|
||
|
|
|
||
|
|
pop ebx
|
||
|
|
|
||
|
|
ret 16
|
||
|
|
}
|
||
|
|
|
||
|
|
#undef DVND
|
||
|
|
#undef DVSR
|
||
|
|
}
|
||
|
|
|
||
|
|
__declspec(naked) void _aullshr()
|
||
|
|
{
|
||
|
|
__asm
|
||
|
|
{
|
||
|
|
cmp cl,64
|
||
|
|
jae short RETZERO
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 0 and 31 bits
|
||
|
|
;
|
||
|
|
cmp cl, 32
|
||
|
|
jae short MORE32
|
||
|
|
shrd eax,edx,cl
|
||
|
|
shr edx,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; Handle shifts of between 32 and 63 bits
|
||
|
|
;
|
||
|
|
MORE32:
|
||
|
|
mov eax,edx
|
||
|
|
xor edx,edx
|
||
|
|
and cl,31
|
||
|
|
shr eax,cl
|
||
|
|
ret
|
||
|
|
|
||
|
|
;
|
||
|
|
; return 0 in edx:eax
|
||
|
|
;
|
||
|
|
RETZERO:
|
||
|
|
xor eax,eax
|
||
|
|
xor edx,edx
|
||
|
|
ret
|
||
|
|
}
|
||
|
|
}
|
||
|
|
}
|
||
|
|
|
||
|
|
#undef CRT_LOWORD
|
||
|
|
#undef CRT_HIWORD
|
||
|
|
|
||
|
|
#endif
|