几个汇编语言的问题
1.MOV CL,20H[BX][SI]；结果(CL)=(DS:[BX+SI+20H])=(091DH:0056H)=(09226H)=0F6H
2.MOV [BP][DI],CX]；结果是将CX的值5678H存入内存(SS:[BP+DI])=(1E4AH:0056H)=(1E4F6H)中，即结果是(1E4F6H)=5678H
3.LEA BX,20H[BX][SI])；结果是将内存偏移地址[BX+SI+20H]=0056H送给BX，即结果(BX)=0056H
MOV AX,2[BX]；结果是(AX)=(DS:[BX+2])=(091DH:0058H)=(09228H)=1E40H
4.LDS SI,[BX][DI]；结果是将内存双字单元(DS:[BX+DI])=(091DH:0056H)=(09226H)中的内容分别送给SI和DS,即结果为(SI)=00F6H,(DS)=1E40H
MOV [SI],BX；结果是将BX值0024H送给内存单元(DS:[SI])=(1E40H:00F6H)=(1E4F6H)中
即最终结果为(1E4F6H)=0024H
5.XCHG CX,32H[BX]；结果是将CX的值5678H和内存单元(DS:[BX+32H])=(091DH:0056)=(09226H)中的内容00F6H互换，互换后结果为(CX)=00F6H,(09226H)=5678H
XCHG 20H[BX][SI],AX；结果是将AX的值1234H与内存单元(DS:[BX+SI+20H])=(091DH:0056H)=(09226H)中的内容5678H互换，互换后结果为(AX)=5678H,(09226H)=1234H
汇编编程问题
1
DATA SEGMENT;
HH DB'PLEASE INPUT NUMBERS!'
STR1 DB 0DH,0AH,'$'
DATA ENDS
STACK SEGMENT STACK
DW 100 DUP(?)
STACK ENDS
CODE SEGMENT
ASSUME CS:CODE,SS:STACK,DS:DATA
START: MOV AX,DATA
MOV DS,AX
MOV DX,OFFSET HH
MOV AH,09H
INT 21H
MOV AH,01H
INT 21H
MOV DL,AL
MOV AH,0AH
INT 21H
MOV AH,4CH
INT 21H
CODE ENDS
END START
汇编编程问题
assume cs:code,ds:data,ss:stack
data segment
da1 db 5,?,0,0,0,0,0
data ends
stack segment stack
dw 10 dup (0)
stack ends
code segment
start:
mov ax,data
mov ds,ax
lea dx,da1
mov ah,0ah
int 21h
lea bx,da1+2
mov si,0
goon:
cmp si,3
ja over1
mov al,[bx+si]
cmp al,'A'
jnb other1
sub al,30h
jmp over2
other1:
cmp al,'a'
jnb other2
sub al,'A'
add al,10
jmp over2
other2:
sub al,'a'
add al,10
over2:
mov [bx+si],al
inc si
jmp goon
over1:
mov dh,[bx]
mov cl,4
sal dh,cl
add dh,[bx+1]
mov dl,[bx+2]
sal dl,cl
add dl,[bx+3]
mov ax,4c00h
int 21h
code ends
end start
一个汇编程序的问题
左移一位相当于乘以2，右移一位相当于除以2