执行后ZF和CF的值置为1
执行后ZF的值置为1
有些指令的执行结果会影响到一些标志位,如指令add/sub的执行会影响到CF, PF, AF, ZF, SF, OF; 有些,则不影响,如mov, push, pop, inc, dec。
inc不改变标志位CF,而add则改变CF
assume cs:code, ds:data data segment x dw 1020h, 2240h, 9522h, 5060h, 3359h, 6652h, 2530h, 7031h y dw 3210h, 5510h, 6066h, 5121h, 8801h, 6210h, 7119h, 3912h data ends code segment start: mov ax, data mov ds, ax mov si, offset x mov di, offset y call add128 mov ah, 4ch int 21h add128: push ax push cx push si push di sub ax, ax mov cx, 8 s: mov ax, [si] adc ax, [di] mov [si], ax inc si inc si inc di inc di loop s pop di pop si pop cx pop ax ret code ends end start
修改前代码测试结果:
修改后代码测试结果:
不能替换,因为add指令会改变CF标志位
数据段x的数据与数据段y的数据做128位加之后,结果保存在数据段x的原有数据的位置上。
运行并测试程序:
①从键盘读取一窜字符,直到读到“#”为止
②实现换行的功能
③将输入的字符串输出到屏幕上
task3.asm:
assume cs:code, ds:data data segment x dw 91, 792, 8536, 65521, 2021 len equ $ - x data ends code segment start: mov ax, data mov ds, ax mov di, offset x mov cx,len mov bx ,10 s1: mov ax,word ptr ds:[di] call printNumber call printSpace inc di inc di dec cx loop s1 mov ah ,4ch int 21h printNumber: push cx mov cx,0 s0: sub ax, 0 jz s3 mov dx,0 div bx push dx inc cx jmp s0 s3:pop dx or dl , 30h mov ah , 2 int 21h loop s3 pop cx ret printSpace: mov ah , 2 mov dl , ' ' int 21h ret code ends end start
运行结果:
代码实现:
assume cs:code, ds:data data segment str db "assembly language, it's not difficult but tedious" len equ $ - str data ends code segment start: mov ax, data mov ds, ax mov si, offset str mov cx,len s1: call strupr mov bl,al mov dl , bl mov ah,2 int 21h inc si loop s1 mov ah ,4ch int 21h strupr: mov al,ds:[si] cmp al,'a' jb next cmp al,'z' ja next sub al,20h next: ret code ends end start
运行并测试程序:
转换成功!
运行并测试代码:
代码功能:从键盘读取一个字符,如果字符是‘7’,则在屏幕输出str1处的字符串”Yes“,反之输出位于str2的字符串”No“
Task6_1.asm
assume cs:code code segment start: ; 42 interrupt routine install code mov ax, cs mov ds, ax mov si, offset int42 ; set ds:si mov ax, 0 mov es, ax mov di, 200h ; set es:di mov cx, offset int42_end - offset int42 cld rep movsb ; set IVT(Interrupt Vector Table) mov ax, 0 mov es, ax mov word ptr es:[42*4], 200h mov word ptr es:[42*4+2], 0 mov ah, 4ch int 21h int42: jmp short int42_start str db "welcome to 2049!" len equ $ - str ; display string "welcome to 2049!" int42_start: mov ax, cs mov ds, ax mov si, 202h mov ax, 0b800h mov es, ax mov di, 24*160 + 32*2 mov cx, len s: mov al, [si] mov es:[di], al mov byte ptr es:[di+1], 2 inc si add di, 2 loop s iret int42_end: nop code ends end start
Task6_2.asm
assume cs:code code segment start: int 42 mov ah, 4ch int 21h code ends end start
运行测试: