1:無符號乘法指令MUL

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
val1 WORD 2000h
val2 WORD 0100h
.code
main PROC;8位乘法mov al,5h  ;被乘數mov bl,10hmul bl     ;AX = 0050h ,CF,OF = 0;16位乘法mov ax,val1mul val2   ;DX:AX = 00200000h ，CF ,OF= 1 高部分不為零0了，CF,OF=1mov eax,12345hmov ebx,1000hmul ebx   ;EDX:EAX = 0000000012345000 ,CF,OF = 0INVOKE ExitProcess,0
main ENDP
END main

2:有符號乘法指令IMUL 單個操作數情況

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.code
main PROCmov al,48mov bl,4imul bl    ;AX = 00C0h ,OF=1   因為AH不是AL的符號擴展，因此OF =1mov al,-4mov bl,4imul bl  ;AX = FFF0h,OF =0 因為AH是AL的擴展，OF清零mov ax,48mov bx,4imul bx   ;DX:AX = 000000C0h,OF = 0 ;因為DX是AX的符號擴展，因此OF=0mov eax,+4823424mov ebx,-423imul ebx  ;EDX:EAX = FFFFFFFF86635D80h, OF = 0,因為EDX是EAX的符號擴展,OF = 0INVOKE ExitProcess,0
main ENDP
END main

3:有符號乘法指令IMUL 兩個操作數情況， 三個操作數情況

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
word1 SWORD 4
dword1 SDWORD 4.code
main PROCmov ax,-16     ;AX = -16mov bx,2       ;BX = 2imul bx,ax     ;BX = -32imul bx,2      ;BX = -64imul bx,word1  ;BX = -256mov eax,-16    ;EAX = -16mov ebx,2      ;EBX = 2imul ebx,eax   ;EBX = -32imul ebx,2     ;EBX = -64imul ebx,dword1;EBX = -256;因為雙操作數與三操作IMUL指令的目的操作數大小與乘積相同;因此會出現符號溢出mov ax,-32000imul ax,2     ;OF = 1 因為-64000不適合16位目的操作數，換成32就不會;三操作數imul bx,word1,-16  ;BX = word1 * -16imul ebx,dword1,-16 ;EBX = dword1 * -16imul ebx,dword1,-2000000000 ;符號溢出INVOKE ExitProcess,0
main ENDP
END main

4:對比MUL乘法指令與SHL移位方式的執行時間對比 （在現在大部分處理器執行時間都差不多了）

include Irvine32.inc.stack 4096
ExitProcee PROTO,dwExitCode:DWORDLOOP_COUNT = 0FFFFFFFFh
.data
intval DWORD 5
startTime DWORD ?.code
main PROCcall GetMseconds  ;獲取開始時間mov startTime,eaxmov eax,intvalcall mult_by_shiftingcall GetMsecondssub eax,startTimecall WriteDeccall Crlfcall GetMseconds  ;獲取開始時間mov startTime,eaxmov eax,intvalcall mult_by_MULcall GetMsecondssub eax,startTimecall WriteDecINVOKE ExitProcess,0
main ENDPmult_by_shifting PROCmov ecx, LOOP_COUNT
L1:push eax  ;保存原始EAXmov ebx,eaxshl eax,5shl ebx,2add eax,ebxpop eax  ;恢復EAXloop L1retmult_by_shifting ENDPmult_by_MUL PROCmov ecx,LOOP_COUNT
L1:push eax  ;保存原始EAXmov ebx,36mul ebxpop eax  ;恢復EAXloop L1ret
mult_by_MUL ENDPEND main

5:DIV 無符號除法指令

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
dividend QWORD 0000000800300020h
divisor  DWORD 00000010h.code
main PROC;8位 AL商，AH余數mov ax,0083h  ;被除數mov bl,2      ;除數div bl        ;AL = 41h  ，AH =01h;16位，AX商，DX余數mov dx,0      ;清除被除數高16位mov ax,8003h  mov cx,100hdiv cx        ;AX=0080h  DX=0003h;32位 EAX商，EDX 余數mov edx,DWORD PTR dividend +4  ;高雙字mov eax,DWORD PTR dividend     ;低雙字div divisor       ;EAX = 08003000h  EDX = 00000020hINVOKE ExitProcess,0
main ENDP
END main

6:符號位擴展指令，CBW,CWD,CDQ

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
byteVal SBYTE -101
wordVal SWORD -101
dwordVal SDWORD -101.code
main PROCmov al,byteVal  ;AL = 9Bhcbw             ;AX = FF9Bhmov ax,wordVal  ;AX = FF9Bhcwd             ;DX:AX = FFFFFFF9Bhmov eax,dwordVal ;EAX = FFFFFF9Bhcdq              ;EDX:EAX = FFFFFFFFFFFFFF9BhINVOKE ExitProcess,0
main ENDP
END main

7:IDIV 有符號除法指令

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
byteVal SBYTE -48
wordVal SWORD -5000
dwordVal SDWORD +50000.code
main PROCmov al,byteVal  ;被除數低字節cbw             ;擴展到AHmov bl,+5idiv bl   ;AL = -9 ,AH =-3mov ax,wordVal  ;被除數低字cwd             ;擴展到DXmov bx,+256 idiv bx         ;商AX = -19  ,余DX = -136mov eax,dwordValcdqmov ebx,-256idiv ebx   ;EAX = -195  EDX = +80INVOKE ExitProcess,0
main ENDP
END main

8:進行除法時，檢查除數不能為0

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
dividend WORD 0083h
divisor BYTE 2.code
main PROCmov ax,dividendmov bl,divisorcmp bl,0je NoDivideZero  ;為零就不進行除div blNoDivideZero:INVOKE ExitProcess,0
main ENDP
END main

9:計算算式 var4 = (var1 + var2)*var3

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
var1 DWORD 10
var2 DWORD 20
var3 DWORD 30
var4 DWORD ?.code
main PROC;var4 = (var1 + var2)*var3mov eax,var1add eax,var2mul var3jc tooBig   ;判斷EAX是否放得下mov var4,eaxjmp next
tooBig:
next:INVOKE ExitProcess,0
main ENDP
END main

10:計算算式 var4 = (var1 * -5) / (-var2 % var3)

.386
.model flat,stdcall.stack 4096
ExitProcess PROTO,dwExitCode:DWORD.data
var1 DWORD 10
var2 DWORD 20
var3 DWORD 30
var4 DWORD ?.code
main PROC;var4 = (var1 * -5) / (-var2 % var3)mov eax,var2neg eaxcdqidiv var3  ;EDX=余數mov ebx,edx mov eax,-5imul var1  ;EDX:EAX = 左邊表達式結果idiv ebxmov var4,eaxINVOKE ExitProcess,0
main ENDP
END main