（汇编源代码 ）侦测CPU型号
_PROCESSOR DETECTION SCHEMES_
by Richard C. Leinecker
[LISTING ONE]
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;; Detect the Processor Type -- by Richard C. Leinecker ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
_PTEXT SEGMENT PARA PUBLIC 'CODE'
ASSUME CS:_PTEXT, DS:_PTEXT
public _Processor
; This routine returns the processor type as an integer value.
; C Prototype
; int Processor( void );
; Returns: 0 == 8088, 1 == 8086, 2 == 80286, 3 == 80386, 4 == 80486
; Code assumes es, ax, bx, cx, and dx can be altered. If their contents
; must be preserved, then you'll have to push and pop them.
_Processor proc far
push bp ; Preserve bp
mov bp,sp ; bp = sp
push ds ; Preserve ds
push di ; and di
mov ax,cs ; Point ds
mov ds,ax ; to _PTEXT
call IsItAn8088 ; Returns 0 or 2
cmp al,2 ; 2 = 286 or better
jge AtLeast286 ; Go to 286 and above code
call IsItAn8086 ; Returns 0 or 1
jmp short ExitProcessor ; Jump to exit code
AtLeast286:
call IsItA286 ; See if it's a 286
cmp al,3 ; 4 and above for 386 and up
jl short ExitProcessor ; Jump to exit code
AtLeast386:
call IsItA386 ; See if it's a 386
ExitProcessor:
pop di ; Restore di,
pop ds ; ds,
pop bp ; and bp
ret ; Return to caller
_Processor endp
; Is It an 8088?
; Returns ax==0 for 8088/8086, ax==2 for 80286 and above
IsItAn8088 proc
pushf ; Preserve the flags
pushf ; Push the flags so
pop bx ; we can pop them into bx
and bx,00fffh ; Mask off bits 12-15
push bx ; and put it back on the stack
popf ; Pop value back into the flags
pushf ; Push it back. 8088/8086 will
; have bits 12-15 set
pop bx ; Get value into bx
and bx,0f000h ; Mask all but bits 12-15
sub ax,ax ; Set ax to 8088 value
cmp bx,0f000h ; See if the bits are still set
je Not286 ; Jump if not
mov al,2 ; Set for 286 and above
Not286:
popf ; Restore the flags
ret ; Return to caller
IsItAn8088 endp
; Is It an 8086?
; Returns ax==0 for 8088, ax==1 for 8086
; Code takes advantage of the 8088's 4-byte prefetch queues and 8086's
; 6-byte prefetch queues. By self-modifying the code at a location exactly 5
; bytes away from IP, and determining if the modification took effect,
; you can differentiate between 8088s and 8086s.
IsItAn8086 proc
mov ax,cs ; es == code segment
mov es,ax
std ; Cause stosb to count backwards
mov dx,1 ; dx is flag and we'll start at 1
mov di,OFFSET EndLabel ; di==offset of code tail to modify
mov al,090h ; al==nop instruction opcode
mov cx,3 ; Set for 3 repetitions
REP stosb ; Store the bytes
cld ; Clear the direction flag
nop ; Three nops in a row
nop ; provide dummy instructions
nop
dec dx ; Decrement flag (only with 8088)
nop ; dummy instruction--6 bytes
EndLabel:
nop
mov ax,dx ; Store the flag in ax
ret ; Back to caller
IsItAn8086 endp
; Is It an 80286?
; Determines whether processor is a 286 or higher. Going into subroutine ax = 2
; If the processor is a 386 or higher, ax will be 3 before returning. The
; method is to set ax to 7000h which represent the 386/486 NT and IOPL bits
; This value is pushed onto the stack and popped into the flags (with popf).
; The flags are then pushed back onto the stack (with pushf). Only a 386 or 486
; will keep the 7000h bits set. If it's a 286, those bits aren't defined and
; when the flags are pushed onto stack these bits will be 0. Now, when ax is
; popped these bits can be checked. If they're set, we have a 386 or 486.
IsItA286 proc
pushf ; Preserve the flags
mov ax,7000h ; Set the NT and IOPL flag
; bits only available for
; 386 processors and above
push ax ; push ax so we can pop 7000h
; into the flag register
popf ; pop 7000h off of the stack
pushf ; push the flags back on
pop ax ; get the pushed flags
; into ax
and ah,70h ; see if the NT and IOPL
; flags are still set
mov ax,2 ; set ax to the 286 value
jz YesItIsA286 ; If NT and IOPL not set
; it's a 286
inc ax ; ax now is 4 to indicate
; 386 or higher
YesItIsA286:
popf ; Restore the flags
ret ; Return to caller
IsItA286 endp
; Is It an 80386 or 80486?
; Determines whether processor is a 386 or higher. Going into subroutine ax=3
; If the processor is a 486, ax will be 4 before leaving. The method is to set
; the AC bit of the flags via EAX and the stack. If it stays set, it's a 486.
IsItA386 proc
mov di,ax ; Store the processor value
mov bx,sp ; Save sp
and sp,0fffch ; Prevent alignment fault
.386
pushfd ; Preserve the flags
pushfd ; Push so we can get flags
pop eax ; Get flags into eax
or eax,40000h ; Set the AC bit
push eax ; Push back on the stack
popfd ; Get the value into flags
pushfd ; Put the value back on stack
pop eax ; Get value into eax
test eax,40000h ; See if the bit is set
jz YesItIsA386 ; If not we have a 386
add di,2 ; Increment to indicate 486
YesItIsA386:
popfd ; Restore the flags
.8086
mov sp,bx ; Restore sp
mov ax,di ; Put processor value into ax
ret ; Back to caller
IsItA386 endp
_PTEXT ENDS
END
[LISTING TWO]
.586
; Pentium detect routine. Call only after verifying processor is an i486 or
; later. Returns 4 if on i486, 5 if Pentium, 6 or greater for future
; Intel processors.
EF_IDequ200000h ; ID bit in EFLAGS register
Pentium proc near
; Check for Pentium or later by attempting to toggle the ID bit in EFLAGS reg;
; if we can't, it's an i486.
pushfd ; get current flags
popeax ;
movebx,eax ;
xoreax,EF_ID; attempt to toggle ID bit
pusheax ;
popfd ;
pushfd ; get new EFLAGS
popeax ;
pushebx ; restore original flags
popfd ;
andeax,EF_ID; if we couldn't toggle ID, it's an i486
andebx,EF_ID ;
cmpeax,ebx ;
jeshort Is486 ;
; It's a Pentium or later. Use CPUID to get processor family.
moveax,1 ; get processor info form of CPUID
cpuid ;
shrax,8 ; get Family field; 5 means Pentium.
andax,0Fh ;
ret
Is486:
movax,4 ; return 4 for i486
ret
pentium endp
[LISTING THREE]
#pragma inline
main()
{
long start1;
long end1;
long start2;
long end2;
start1 = clock();
asm P386
asm mov eax,10000000
asm lea ebx,loop1
asm loop1:
asm dec eax
asm jz loop1e
asm jmp ebx
loop1e:
start2 = end1 = clock();
asm mov eax,10000000
asm lea ebx,loop2
asm nop
asm loop2:
asm dec eax
asm jz loop2e
asm jmp ebx
loop2e:
end2 = clock();
printf("misaligned loop time = %d, aligned loop time =%d
",
(int)(end1-start1), (int)(end2-start2));
return;
}
by Richard C. Leinecker
[LISTING ONE]
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Detect the Processor Type -- by Richard C. Leinecker ;;
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
_PTEXT SEGMENT PARA PUBLIC 'CODE'
ASSUME CS:_PTEXT, DS:_PTEXT
public _Processor
; This routine returns the processor type as an integer value.
; C Prototype
; int Processor( void );
; Returns: 0 == 8088, 1 == 8086, 2 == 80286, 3 == 80386, 4 == 80486
; Code assumes es, ax, bx, cx, and dx can be altered. If their contents
; must be preserved, then you'll have to push and pop them.
_Processor proc far
push bp ; Preserve bp
mov bp,sp ; bp = sp
push ds ; Preserve ds
push di ; and di
mov ax,cs ; Point ds
mov ds,ax ; to _PTEXT
call IsItAn8088 ; Returns 0 or 2
cmp al,2 ; 2 = 286 or better
jge AtLeast286 ; Go to 286 and above code
call IsItAn8086 ; Returns 0 or 1
jmp short ExitProcessor ; Jump to exit code
AtLeast286:
call IsItA286 ; See if it's a 286
cmp al,3 ; 4 and above for 386 and up
jl short ExitProcessor ; Jump to exit code
AtLeast386:
call IsItA386 ; See if it's a 386
ExitProcessor:
pop di ; Restore di,
pop ds ; ds,
pop bp ; and bp
ret ; Return to caller
_Processor endp
; Is It an 8088?
; Returns ax==0 for 8088/8086, ax==2 for 80286 and above
IsItAn8088 proc
pushf ; Preserve the flags
pushf ; Push the flags so
pop bx ; we can pop them into bx
and bx,00fffh ; Mask off bits 12-15
push bx ; and put it back on the stack
popf ; Pop value back into the flags
pushf ; Push it back. 8088/8086 will
; have bits 12-15 set
pop bx ; Get value into bx
and bx,0f000h ; Mask all but bits 12-15
sub ax,ax ; Set ax to 8088 value
cmp bx,0f000h ; See if the bits are still set
je Not286 ; Jump if not
mov al,2 ; Set for 286 and above
Not286:
popf ; Restore the flags
ret ; Return to caller
IsItAn8088 endp
; Is It an 8086?
; Returns ax==0 for 8088, ax==1 for 8086
; Code takes advantage of the 8088's 4-byte prefetch queues and 8086's
; 6-byte prefetch queues. By self-modifying the code at a location exactly 5
; bytes away from IP, and determining if the modification took effect,
; you can differentiate between 8088s and 8086s.
IsItAn8086 proc
mov ax,cs ; es == code segment
mov es,ax
std ; Cause stosb to count backwards
mov dx,1 ; dx is flag and we'll start at 1
mov di,OFFSET EndLabel ; di==offset of code tail to modify
mov al,090h ; al==nop instruction opcode
mov cx,3 ; Set for 3 repetitions
REP stosb ; Store the bytes
cld ; Clear the direction flag
nop ; Three nops in a row
nop ; provide dummy instructions
nop
dec dx ; Decrement flag (only with 8088)
nop ; dummy instruction--6 bytes
EndLabel:
nop
mov ax,dx ; Store the flag in ax
ret ; Back to caller
IsItAn8086 endp
; Is It an 80286?
; Determines whether processor is a 286 or higher. Going into subroutine ax = 2
; If the processor is a 386 or higher, ax will be 3 before returning. The
; method is to set ax to 7000h which represent the 386/486 NT and IOPL bits
; This value is pushed onto the stack and popped into the flags (with popf).
; The flags are then pushed back onto the stack (with pushf). Only a 386 or 486
; will keep the 7000h bits set. If it's a 286, those bits aren't defined and
; when the flags are pushed onto stack these bits will be 0. Now, when ax is
; popped these bits can be checked. If they're set, we have a 386 or 486.
IsItA286 proc
pushf ; Preserve the flags
mov ax,7000h ; Set the NT and IOPL flag
; bits only available for
; 386 processors and above
push ax ; push ax so we can pop 7000h
; into the flag register
popf ; pop 7000h off of the stack
pushf ; push the flags back on
pop ax ; get the pushed flags
; into ax
and ah,70h ; see if the NT and IOPL
; flags are still set
mov ax,2 ; set ax to the 286 value
jz YesItIsA286 ; If NT and IOPL not set
; it's a 286
inc ax ; ax now is 4 to indicate
; 386 or higher
YesItIsA286:
popf ; Restore the flags
ret ; Return to caller
IsItA286 endp
; Is It an 80386 or 80486?
; Determines whether processor is a 386 or higher. Going into subroutine ax=3
; If the processor is a 486, ax will be 4 before leaving. The method is to set
; the AC bit of the flags via EAX and the stack. If it stays set, it's a 486.
IsItA386 proc
mov di,ax ; Store the processor value
mov bx,sp ; Save sp
and sp,0fffch ; Prevent alignment fault
.386
pushfd ; Preserve the flags
pushfd ; Push so we can get flags
pop eax ; Get flags into eax
or eax,40000h ; Set the AC bit
push eax ; Push back on the stack
popfd ; Get the value into flags
pushfd ; Put the value back on stack
pop eax ; Get value into eax
test eax,40000h ; See if the bit is set
jz YesItIsA386 ; If not we have a 386
add di,2 ; Increment to indicate 486
YesItIsA386:
popfd ; Restore the flags
.8086
mov sp,bx ; Restore sp
mov ax,di ; Put processor value into ax
ret ; Back to caller
IsItA386 endp
_PTEXT ENDS
END
[LISTING TWO]
.586
; Pentium detect routine. Call only after verifying processor is an i486 or
; later. Returns 4 if on i486, 5 if Pentium, 6 or greater for future
; Intel processors.
EF_IDequ200000h ; ID bit in EFLAGS register
Pentium proc near
; Check for Pentium or later by attempting to toggle the ID bit in EFLAGS reg;
; if we can't, it's an i486.
pushfd ; get current flags
popeax ;
movebx,eax ;
xoreax,EF_ID; attempt to toggle ID bit
pusheax ;
popfd ;
pushfd ; get new EFLAGS
popeax ;
pushebx ; restore original flags
popfd ;
andeax,EF_ID; if we couldn't toggle ID, it's an i486
andebx,EF_ID ;
cmpeax,ebx ;
jeshort Is486 ;
; It's a Pentium or later. Use CPUID to get processor family.
moveax,1 ; get processor info form of CPUID
cpuid ;
shrax,8 ; get Family field; 5 means Pentium.
andax,0Fh ;
ret
Is486:
movax,4 ; return 4 for i486
ret
pentium endp
[LISTING THREE]
#pragma inline
main()
{
long start1;
long end1;
long start2;
long end2;
start1 = clock();
asm P386
asm mov eax,10000000
asm lea ebx,loop1
asm loop1:
asm dec eax
asm jz loop1e
asm jmp ebx
loop1e:
start2 = end1 = clock();
asm mov eax,10000000
asm lea ebx,loop2
asm nop
asm loop2:
asm dec eax
asm jz loop2e
asm jmp ebx
loop2e:
end2 = clock();
printf("misaligned loop time = %d, aligned loop time =%d
",
(int)(end1-start1), (int)(end2-start2));
return;
}
