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Visual C/C++ CRT functionality
extern "C"
{
int _fltused;
#ifdef _M_IX86 // following functions are needed only for 32-bit architecture
__declspec(naked) void _ftol2()
{
__asm
{
fistp qword ptr [esp-8]
mov edx,[esp-4]
mov eax,[esp-8]
ret
}
}
__declspec(naked) void _ftol2_sse()
{
__asm
{
fistp dword ptr [esp-4]
mov eax,[esp-4]
ret
}
}
#if 0 // these functions are needed for SSE code for 32-bit arch, TODO: implement them
__declspec(naked) void _dtol3()
{
__asm
{
}
}
__declspec(naked) void _dtoui3()
{
__asm
{
}
}
__declspec(naked) void _dtoul3()
{
__asm
{
}
}
__declspec(naked) void _ftol3()
{
__asm
{
}
}
__declspec(naked) void _ftoui3()
{
__asm
{
}
}
__declspec(naked) void _ftoul3()
{
__asm
{
}
}
__declspec(naked) void _ltod3()
{
__asm
{
}
}
__declspec(naked) void _ultod3()
{
__asm
{
}
}
#endif
#endif
}
#ifdef _M_IX86 // use this file only for 32-bit architecture
#define CRT_LOWORD(x) dword ptr [x+0]
#define CRT_HIWORD(x) dword ptr [x+4]
extern "C"
{
__declspec(naked) void _alldiv()
{
#define DVND esp + 16 // stack address of dividend (a)
#define DVSR esp + 24 // stack address of divisor (b)
__asm
{
push edi
push esi
push ebx
; 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
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
inc edi ; complement the result sign flag
mov edx,CRT_LOWORD(DVSR) ; lo word of a
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 ; eax <- 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 ; eax <- low order bits of quotient
mov edx,ebx ; edx:eax <- quotient
jmp short L4 ; set sign, 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
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 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 one (1) from the quotient.
;
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:
dec esi ; subtract 1 from quotient
L7:
xor edx,edx ; edx:eax <- quotient
mov eax,esi
;
; Just the cleanup left to do. edx:eax contains the quotient. Set the sign
; according to the save value, cleanup the stack, and return.
;
L4:
dec edi ; check to see if result is negative
jnz short L8 ; if EDI == 0, result should be negative
neg edx ; otherwise, negate the result
neg eax
sbb edx,0
;
; Restore the saved registers and return.
;
L8:
pop ebx
pop esi
pop edi
ret 16
}
#undef DVND
#undef DVSR
}
__declspec(naked) void _alldvrm()
{
#define DVND esp + 16 // stack address of dividend (a)
#define DVSR esp + 24 // stack address of divisor (b)
__asm
{
push edi
push esi
push ebp
; Determine sign of the quotient (edi = 0 if result is positive, non-zero
; otherwise) and make operands positive.
; Sign of the remainder is kept in ebp.
xor edi,edi ; result sign assumed positive
xor ebp,ebp ; 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
inc ebp ; complement result sign flag
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
inc edi ; complement the result sign flag
mov edx,CRT_LOWORD(DVSR) ; lo word of a
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 ; eax <- 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 ; eax <- 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 L4 ; complete remainder calculation
;
; 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
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 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 one (1) from the quotient.
;
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:
dec esi ; subtract 1 from quotient
sub eax,CRT_LOWORD(DVSR) ; subtract divisor from result
sbb edx,CRT_HIWORD(DVSR)
L7:
xor ebx,ebx ; ebx:esi <- quotient
L4:
;
; 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 ebp ; check result sign flag
jns short L9 ; result is ok, set up the quotient
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.
;
L9:
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. Set the sign
; according to the save value, cleanup the stack, and return.
;
dec edi ; check to see if result is negative
jnz short L8 ; if EDI == 0, result should be negative
neg edx ; otherwise, negate the result
neg eax
sbb edx,0
;
; Restore the saved registers and return.
;
L8:
pop ebp
pop esi
pop edi
ret 16
}
#undef DVND
#undef DVSR
}
__declspec(naked) void _allmul()
{
#define A esp + 8 // stack address of a
#define B esp + 16 // stack address of b
__asm
{
push ebx
mov eax,CRT_HIWORD(A)
mov ecx,CRT_LOWORD(B)
mul ecx ;eax has AHI, ecx has BLO, so AHI * BLO
mov ebx,eax ;save result
mov eax,CRT_LOWORD(A)
mul CRT_HIWORD(B) ;ALO * BHI
add ebx,eax ;ebx = ((ALO * BHI) + (AHI * BLO))
mov eax,CRT_LOWORD(A) ;ecx = BLO
mul ecx ;so edx:eax = ALO*BLO
add edx,ebx ;now edx has all the LO*HI stuff
pop ebx
ret 16 ; callee restores the stack
}
#undef A
#undef B
}
__declspec(naked) void _allrem()
{
#define DVND esp + 12 // stack address of dividend (a)
#define DVSR esp + 20 // stack address of divisor (b)
__asm
{
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
extern "C"
{
#pragma function(memset)
void *memset(void *dest, int c, size_t count)
{
char *bytes = (char *)dest;
while (count--)
{
*bytes++ = (char)c;
}
return dest;
}
#pragma function(memcpy)
void *memcpy(void *dest, const void *src, size_t count)
{
char *dest8 = (char *)dest;
const char *src8 = (const char *)src;
while (count--)
{
*dest8++ = *src8++;
}
return dest;
}
}
extern "C"
{
#if _M_IX86
EXCEPTION_DISPOSITION
_except_handler3(
struct _EXCEPTION_RECORD* ExceptionRecord,
void* EstablisherFrame,
struct _CONTEXT* ContextRecord,
void* DispatcherContext)
{
typedef EXCEPTION_DISPOSITION Function(struct _EXCEPTION_RECORD*, void*, struct _CONTEXT*, void*);
static Function* FunctionPtr;
if (!FunctionPtr)
{
HMODULE Library = LoadLibraryA("msvcrt.dll");
FunctionPtr = (Function*)GetProcAddress(Library, "_except_handler3");
}
return FunctionPtr(ExceptionRecord, EstablisherFrame, ContextRecord, DispatcherContext);
}
UINT_PTR __security_cookie = 0xBB40E64E;
extern PVOID __safe_se_handler_table[];
extern BYTE __safe_se_handler_count;
typedef struct {
DWORD Size;
DWORD TimeDateStamp;
WORD MajorVersion;
WORD MinorVersion;
DWORD GlobalFlagsClear;
DWORD GlobalFlagsSet;
DWORD CriticalSectionDefaultTimeout;
DWORD DeCommitFreeBlockThreshold;
DWORD DeCommitTotalFreeThreshold;
DWORD LockPrefixTable;
DWORD MaximumAllocationSize;
DWORD VirtualMemoryThreshold;
DWORD ProcessHeapFlags;
DWORD ProcessAffinityMask;
WORD CSDVersion;
WORD Reserved1;
DWORD EditList;
PUINT_PTR SecurityCookie;
PVOID *SEHandlerTable;
DWORD SEHandlerCount;
} IMAGE_LOAD_CONFIG_DIRECTORY32_2;
const
IMAGE_LOAD_CONFIG_DIRECTORY32_2 _load_config_used = {
sizeof(IMAGE_LOAD_CONFIG_DIRECTORY32_2),
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
0,
&__security_cookie,
__safe_se_handler_table,
(DWORD)(DWORD_PTR) &__safe_se_handler_count
};
#elif _M_AMD64
EXCEPTION_DISPOSITION
__C_specific_handler(
struct _EXCEPTION_RECORD* ExceptionRecord,
void* EstablisherFrame,
struct _CONTEXT* ContextRecord,
struct _DISPATCHER_CONTEXT* DispatcherContext)
{
typedef EXCEPTION_DISPOSITION Function(struct _EXCEPTION_RECORD*, void*, struct _CONTEXT*, _DISPATCHER_CONTEXT*);
static Function* FunctionPtr;
if (!FunctionPtr)
{
HMODULE Library = LoadLibraryA("msvcrt.dll");
FunctionPtr = (Function*)GetProcAddress(Library, "__C_specific_handler");
}
return FunctionPtr(ExceptionRecord, EstablisherFrame, ContextRecord, DispatcherContext);
}
#endif
}
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