汇编语言 实现1.将数据区buf1中的10个数,传送到数据区buf2 2.计算buf1数据的累加和
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1、 将数据区buf1中的10个数,传送到数据区buf2
; multi-segment executable file template.
data segment
buf1 dw 1,2,3,4,5,6,7,8,9,10
buf2 dw ?
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov es, ax
lea si,buf1
lea di,buf2
mov cx,10
rep movsw;拷贝
mov ax, 4c00h ; exit to operating system.
int 21h
ends
end start ; set entry point and stop the assembler.
串传送指令有四种格式
movs DST,SRC
movsb 字节
movsw 字
movsd 双字 (适合386后继机型)
如果不用传送指令
用循环实现
; multi-segment executable file template.
data segment
buf1 dw 1,2,3,4,5,6,7,8,9,10
buf2 dw ?
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov es, ax
lea si,buf1 ;得到buf1的偏移地址
mov di,0 ;定义buf2的偏移地址
mov cx ,10 ;计数器
again:
mov bx,buf1[si]
mov buf2[di],bx
add si,2
add di,2
loop again
mov ax, 4c00h ; exit to operating system.
int 21h
ends
end start ; set entry point and stop the assembler.
运行结束不清楚为什么buf1后面还有多余的数据
如果有知道原因的希望可以告诉我,刚接触汇编。
2.计算buf1数据的累加和
; multi-segment executable file template.
data segment
buf1 dw 1,2,3,4,5,6,7,8,9,10
count dw ?
ends
stack segment
dw 128 dup(0)
ends
code segment
start:
; set segment registers:
mov ax, data
mov ds, ax
mov es, ax
mov si,offset buf1
mov cx,10 ;计数器
mov ax,0 ;求和结果保存
again:
add ax,buf1[si] ;
add si,2 ;因为是字,所以占两个字节
loop again
mov count,ax ;将和传值给count
mov ax, 4c00h ; exit to operating system.
int 21h
ends
end start ; set entry point and stop the assembler.
也是通过循环来实现。
标签:10,ends,buf1,数据,mov,dw,ax,segment 来源: https://blog.csdn.net/JulianBrandt/article/details/88770152