汇编语言例子

文二先生 發表於 2019-7-6 15:01:00

<div style="color: rgba(47, 47, 47, 1); background: left top; font-family: "Helvetica Neue", Helvetica, Tahoma, Arial, "Hiragino Sans GB", STHeiti, "Microsoft YaHei", "微软雅黑", "WenQuanYi Micro Hei", STXihei, "华文细黑", Heiti, "黑体", SimSun, "宋体", Song, sans-serif; font-size: 16px; line-height: 1.33; font-weight: 100; word-wrap: break-word; letter-spacing: 0; word-spacing: 0.25em">
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<h1 class="story_title" style="line-height: 1.6; margin: 0.7em 0; font-family: inherit; font-weight: 500; color: inherit; text-rendering: optimizelegibility; font-size: 3.4rem; text-align: center">汇编语言例子</h1>
<div class="story_tags" style="margin: 0 0 1.2em; text-align: center"><span class="tag blue label" style="font-weight: bold; display: inline-block; vertical-align: baseline; line-height: 1; background: rgba(235, 238, 239, 1); border-color: rgba(235, 238, 239, 1); padding: 0.6em 0.8em; color: rgba(94, 96, 96, 1); text-transform: none; border-radius: 0.2857rem; box-sizing: border-box; font-size: 12px; margin: 0.125em">汇编语言微机原理 8086</div>
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<pre><code class="nohighlight hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em">编写程序，将一个包含有20个数据的数组M分成两个数组：正数数组P和负数数组N，并分别把这两个数组中数据的个数统计出来，依次存放。</code></pre>
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
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<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
mdb 1,2,4,3,5,12,45,66,7,89,55,34,21
db -2,-34,-5,-9,-23,-5,-31 ;注意这里的写法，可以换行db
pdb 20 dup(0)
ndb 20 dup(0)
cp db dup(0)
cn db dup(0)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax;

mov cx,20;数组长度
lea bx,m; 得到数组首地址
mov dl,0;放置cp
mov dh,0;放置cn

lea si,p;
lea di,n

l: mov al,[bx];
cmp al,0 ;
jznext ;等于0 下一个
jgtoN ;正数有符号数比较 greater less 无符号数above below
mov [di],al; ;负数
incdi ;下一个负数地址
incdh ;计数负数
jmp next;
toN: mov [si],al
incsi ;下一个正数地址
incdl ;计数正数
next: inc bx; 处理下一个元素
loopl
mov cp,dh;负数个数;
mov cn,dl;正数个数
code ends
end start</code></pre>
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<pre><code class="nohighlight hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em"> 试编写一个汇编语言程序，求出首地址为DATA的100DH字数组中的最小偶数，并把它存放在AX中。</code></pre>
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
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<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
mydatadw 2,3,45,3,34,2,43,76,45,8,98,67,56,6,78,5,23,33,54
count dw ($-mydata )/ 2
data ends
code segment
 assume: cs:code,ds:data
start:
mov ax,data;
mov ds,ax; 


mov cx,count
mov ax,07ffeh; 最大偶数

lea bx,mydata

l:
 movdx,[bx];取数
 test dx,01h;
 jnznext;不是偶数下一个
 ;是偶数
 cmpdx,ax;
 jge next;大于等于直接下一个
 mov ax,dx ;更新 ax
next: inc bx
 inc bx
 loop l
code ends
end start </code></pre>
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<h2 class="xsj_heading_hash xsj_heading xsj_heading_h2" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 3rem">例3</h2>
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数据段中已定义了一个有n个字数据的数组M，试编写一程序求出M中绝对值最大的数，把它放在数据段的M+2n单元中，并将该数的偏移地址存放在M+2（n+1）单元中（假设n=10）。
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
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<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
m dw 20, 12, 33, -23, -89, 31, 68, 52, 21, 31
count dw ($-m)/2
maxdw dup(0)
maxeadw dup(0)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax

lea bx,m;
mov ax,0ffffh;最小的数字
mov cx,count

l:mov dx,[bx]
cmp dx,0
jgtoN;正数
neg dx
toN:
cmp dx,ax
jle next;

mov ax,dx;更新AX
mov si,bx;存放EA
next:
inc bx;
inc bx;
loop l

mov max,ax
mov maxea,si

mov ah,4ch
int 21h
code ends
end start</code></pre>
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<h2 class="xsj_heading_hash xsj_heading xsj_heading_h2" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 3rem">例4</h2>
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<pre><code class="nohighlight hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em">在首地址为DATA的字数组中，存放了100H个16位补码数，试编写一程序，求出他们的平均值，放在AX寄存器中:并求出数组中有多少个数低于此平均值，将结果放在BX寄存器中</code></pre>
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
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<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
mydata dw 1,2,3,4,5,6,7,8,9,0
count dw ($-mydata)/2
avr dw dup(0)
below_count dw dup(0)
data ends
code segment
assume cs:code,ds:data
start:
mov ax,data
mov ds,ax

lea bx,mydata
mov cx,count;

;计算平均值
;累加 16 位要考虑溢出
xor ax,ax;
xor dx,dx
sum:
add ax,[bx];
adc dx,0 ;处理进位

inc bx;
inc bx;

loop sum;

;除法
mov cx,count;
idiv cx;

mov avr,ax
;
;计数
lea bx,mydata
mov cx,count;
mov dx,avr
l:mov ax,[bx]
cmp ax,dx;
jge next;
inc below_count;
next:
inc bx
inc bx
loop l; 

mov ah,4ch
int 21h
code ends
end start</code></pre>
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<h2 class="xsj_heading_hash xsj_heading xsj_heading_h2" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 3rem">例5</h2>
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<div class="xiaoshujiang_code_container" style="margin: 1em 0; position: relative; padding-bottom: 2em">
<pre><code class="nohighlight hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em">在当前数据段偏移地址为DATAB开始的顺序80个单元中，存放着某班80名同学某门考试的成绩。
要求：
1）编写完整源程序统计大于等于90分、80～89分、70～79分、60～69分、小于60分的人数各为多少，并将结果放在同一数据段以BTRX开始的顺序单元中。
2）编写完整源程序求该班这门课程的平均成绩，并放在同一数据段的LEVT单元。</code></pre>
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
<div class="xiaoshujiang_code_container" style="margin: 1em 0; position: relative; padding-bottom: 2em">
<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
mydatadb 100,99
 db 89,88
 db 79,78
 db 69,68
 db 59,58
count dw $-mydata
btrx db 0
 db 0
 db 0
 db 0
 db 0
levt db 0
data ends
code segment
assume cs:code, ds:data
start:
mov ax,data;
mov ds,ax

lea bx,mydata
mov cx,count
l:
mov al,[bx]; mov ax, 
l90:
cmp al,90
jb l80;
inc btrx
jmp next
l80:
cmp al,80
jb l70;
inc
jmp next
l70:
cmp al,70
jb l60;
inc
jmp next
l60:
cmp al,60
jb l00;
inc
jmp next
l00:
inc
next:
inc bx;
loop l;
;求和关键在于注意溢出 
lea bx,mydata
mov cx,count

xor ax,ax;
sum:
add al,[bx]
adc ah,0
inc bx
loop sum

;除法 
mov cx ,count;
div cl;

mov levt,al

mov ah,4ch;
int 21h;
code ends
endstart</code></pre>
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<div class="xiaoshujiang_code_container" style="margin: 1em 0; position: relative; padding-bottom: 2em">
<pre><code class="nohighlight hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em">编写程序段实现根据AX寄存器中1的位置转向SUB0～SUB15这16个分支程序执行。设AX中最多只能有一个位为1。</code></pre>
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<h3 class="xsj_heading_hash xsj_heading xsj_heading_h3" style="margin: 1em 0 0.6em; font-family: inherit; font-weight: 500; line-height: 1.1; color: inherit; text-rendering: optimizelegibility; font-size: 2.45rem">程序</h3>
<div class="xiaoshujiang_code_container" style="margin: 1em 0; position: relative; padding-bottom: 2em">
<pre><code class="language-x86asm hljs" style="background: rgba(40, 43, 46, 1); color: rgba(169, 183, 198, 1); border-radius: 4px; border: 0; font-size: 90%; overflow: visible !important; word-spacing: initial; display: block; overflow-x: auto; padding: 0.5em" data-info="x86asm">data segment
myfun dw sub0,sub1,sub2 ;关键在于分支子程序放置的位置在code ends 与 end start 之间 
wa dw 0,0
str0 db "howbin$"
str1 db "dcx$"
str2 db "wawaw$"
data ends
stack segment
dw 128dup(0)
stack ends
code segment
assume cs:code,ds:data,ss:stack
start:
mov ax,data;
mov ds,ax;

mov ax,stack
mov ss,ax

mov al,4;
xor bx,bx;

l: shr al,1
jctoSub;
inc bx;
inc bx;
jnzl;
jmp over;
toSub:
callmyfun[bx]; 
over:

mov ah,ch
int 21h

code ends

sub0 proc
lea dx,str0
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub0 endp

sub1 proc
lea dx,str1
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub1 endp

sub2 proc
lea dx,str2
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub2 endp

end start data segment
myfun dw sub0,sub1,sub2 ;关键在于分支子程序放置的位置在code ends 与 end start 之间 
wa dw 0,0
str0 db "howbin$"
str1 db "dcx$"
str2 db "wawaw$"
data ends
stack segment
dw 128dup(0)
stack ends
code segment
assume cs:code,ds:data,ss:stack
start:
mov ax,data;
mov ds,ax;

mov ax,stack
mov ss,ax

mov al,4;
xor bx,bx;

l: shr al,1
jctoSub;
inc bx;
inc bx;
jnzl;
jmp over;
toSub:
callmyfun[bx]; 
over:

mov ah,ch
int 21h

code ends

sub0 proc
lea dx,str0
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub0 endp

sub1 proc
lea dx,str1
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub1 endp

sub2 proc
lea dx,str2
mov ah,9;
int 21h

mov ah, 1
int 21h

ret
sub2 endp

end start</code></pre>
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来源：https://www.cnblogs.com/Howbin/p/11142765.html

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汇编语言例子