【汇编语言】实验二

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我是靠谱客的博主老迟到黑夜，这篇文章主要介绍【汇编语言】实验二，现在分享给大家，希望可以做个参考。

X、Y、Z、V均为字变量，在X、Y、Z、V字单元中存放是16位带符号数。试编写汇编语言程序完成以下功能：
计算表达式值（V–（X*Y+Z-720））/X，将运算结果整数放在SUM1单元，余数放在SUM2单元。
（1）在DOSBOX编译链接成可执行文件后，使用Debug装入内存。
（2）使用U命令反汇编代码，并与源文件比对，需要注意数据段名、变量转入内存后的形式。
（3）分别在G命令执行前后，使用D命令查看各个变量的值。

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
X dw 540
Y dw 1
Z dw -1
V dw 1080
SUM1 dw ?
SUM2 dw ? 
data ends
code segment
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	mov ax,word ptr X
	imul word ptr Y
	mov bx,ax
	mov cx,dx
	mov ax,word ptr Z
	cwd
	add bx,ax
	adc cx,dx
	sub bx,720
	sbb cx,0
	mov ax,word ptr V
	cwd
	sub ax,bx
	sbb dx,cx
	idiv x
	mov word ptr SUM1,ax
	mov word ptr SUM2,dx
	mov ax,4c00h
	int 21h


code ends
end start

在这里插入图片描述

2、BL中的只有一位为0，编写程序测试0所在的位数（从左编号，最左边为第0位），并输出提示信息“The X Bit is 0”（X为0、1、2、3…7），要求使用地址表方法实现。

首先，没有用地址表的方法。

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
str1 db "The $"
str2 db " Bit is 0$"
data ends
code segment
printstr macro str	;宏--用于输出字符串
push ax
push dx
lea dx,str
mov ah,09h
int 21h
pop dx
pop ax
endm
printnum macro num	;宏--用于输出一个十进制的数
push ax
push dx
xor dx,dx
mov dx,30h
add dx,num
mov ah,02h
int 21h
pop dx
pop ax
endm
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	mov bl,11011111b
	xor ax,ax			;记录位数
	xor dx,dx			;判断是否为0
	mov cx,8
again:	sar bl,1
	adc dx,0
	cmp dx,0
	jz done
	inc ax
	xor dx,dx
	loop again
done:	printstr str1
	printnum ax
	printstr str2
	mov ax,4c00h
	int 21h


code ends
end start

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
table dw l0,l1,l2,l3,l4,l5,l6,l7
str0 db 'The 0 Bit is 0$'
str1 db 'The 1 Bit is 0$'
str2 db 'The 2 Bit is 0$'
str3 db 'The 3 Bit is 0$'
str4 db 'The 4 Bit is 0$'
str5 db 'The 5 Bit is 0$'
str6 db 'The 6 Bit is 0$'
str7 db 'The 7 Bit is 0$'
str db 'bl = $'
data ends
code segment
printstr macro str	;宏--用于输出字符串
push ax
push dx
lea dx,str
mov ah,09h
int 21h
pop dx
pop ax
endm
printnum macro num	;宏--用于输出二进制字符
push ax
push cx
push dx
mov cx,8
mov al,num
xor dx,dx
again1:sar al,1
adc dx,30h
push dx
xor dx,dx
loop again1
mov cx,8
continue:pop dx
mov ah,02h
int 21h
loop continue
mov dl,62h
int 21h
mov dl,0dh
int 21h
mov dl,0ah
int 21h
pop dx
pop cx
pop ax
endm
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	mov bl,11011111b
	printstr str
	printnum bl
	xor si,si
again:	sar bl,1
	inc si
	jc again
	dec si
	shl si,1
	jmp table[si]
l0:	mov dx,offset str0
	mov ah,09h
	int 21h
	jmp done
l1:	mov dx,offset str1
	mov ah,09h
	int 21h
	jmp done
l2:	mov dx,offset str2
	mov ah,09h
	int 21h
	jmp done
l3:	mov dx,offset str3
	mov ah,09h
	int 21h
	jmp done
l4:	mov dx,offset str4
	mov ah,09h
	int 21h
	jmp done
l5:	mov dx,offset str5
	mov ah,09h
	int 21h
	jmp done
l6:	mov dx,offset str6
	mov ah,09h
	int 21h
	jmp done
l7:	mov dx,offset str7
	mov ah,09h
	int 21h
done:	mov ax,4c00h
	int 21h

code ends
end start

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
table dw l0,l1,l2,l3,l4,l5,l6,l7
str0 db 'The 0 Bit is 0$'
str1 db 'The 1 Bit is 0$'
str2 db 'The 2 Bit is 0$'
str3 db 'The 3 Bit is 0$'
str4 db 'The 4 Bit is 0$'
str5 db 'The 5 Bit is 0$'
str6 db 'The 6 Bit is 0$'
str7 db 'The 7 Bit is 0$'
str db 'bl = $'
data ends
code segment
printstr macro str	;宏--用于输出字符串
push ax
push dx
lea dx,str
mov ah,09h
int 21h
pop dx
pop ax
endm
printnum macro num	;宏--用于输出二进制字符
push ax
push cx
push dx
mov cx,8
mov al,num
xor dx,dx
again1:sar al,1
adc dx,30h
push dx
xor dx,dx
loop again1
mov cx,8
continue:pop dx
mov ah,02h
int 21h
loop continue
mov dl,62h
int 21h
mov dl,0dh
int 21h
mov dl,0ah
int 21h
pop dx
pop cx
pop ax
endm
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	mov bl,11011111b
	printstr str
	printnum bl
	xor si,si
again:	sar bl,1
	inc si
	jc again
	dec si
	shl si,1
	jmp table[si]
l0:	mov dx,offset str0
	mov ah,09h
	int 21h
	jmp done
l1:	mov dx,offset str1
	mov ah,09h
	int 21h
	jmp done
l2:	mov dx,offset str2
	mov ah,09h
	int 21h
	jmp done
l3:	mov dx,offset str3
	mov ah,09h
	int 21h
	jmp done
l4:	mov dx,offset str4
	mov ah,09h
	int 21h
	jmp done
l5:	mov dx,offset str5
	mov ah,09h
	int 21h
	jmp done
l6:	mov dx,offset str6
	mov ah,09h
	int 21h
	jmp done
l7:	mov dx,offset str7
	mov ah,09h
	int 21h
done:	mov ax,4c00h
	int 21h


code ends
end start

在这里插入图片描述

3、在内存Score缓冲区中存放有100个学生的成绩数据，为无符号字节数。设计程序完成如下功能：根据用户输入的一个2位十进制数，作为查找对象，在该数组中查找，若找到则显示“Y”，若没找到则显示“N” 。

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
Score db 23,34,56,86,95,34,76,79,45,65

data ends
code segment
printChar macro Char	;宏--用于输出字符 'Y' 或‘N’
push ax
push dx
mov dl,Char
mov ah,02h
int 21h
pop dx
pop ax
endm

start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	sub sp,2
	call input
	pop ax
	mov cx,10
	mov si,offset Score
continue:	cmp byte ptr [si],al
	je l
	inc si
	loop continue
	printChar 'N'
	jmp done
l:	printChar 'Y'
	
done:	mov ax,4c00h
	int 21h

input proc	;从键盘输入一个数，0~65535
push bp
mov bp,sp
push ax
push bx
push cx
mov bx,10
mov word ptr[bp+4],0
again:mov ah,01h
int 21h
cmp al,0dh
je next
xor cx,cx
add cl,al
sub cl,30h
mov ax,word ptr [bp+4]
mul bx
add ax,cx
mov word ptr[bp+4],ax
jmp again
next:pop cx
pop bx
pop ax
pop bp
ret
input endp

code ends
end start

在这里插入图片描述
4. 编写一个子程序计算z=f(x,y)=x*y+x-y（x,y,z有符号数字操作数，假设计算结果为16位，可以放在z中）。要求：（1）通过堆栈传送参数x和y；（2）将计算结果按照10进制输出到屏幕上。

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
str1 db 'please input x: $'
str2 db 'please input y: $'
str3 db 'z = $'
x dw ?
y dw ?
z dw ?
data ends
code segment
newline macro	;宏--用于输出回车换行
push ax
push dx
mov dl,0dh
mov ah,02h
int 21h
mov dl,0ah
mov ah,02h
int 21h
pop dx
pop ax
endm
printstr macro str	;宏--用于输出字符串
push ax
push dx
lea dx,str
mov ah,09h
int 21h
pop dx
pop ax
endm
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	;用来输入x,y
	printstr str1
	sub sp,2
	call input
	pop word ptr x
	printstr str2
	sub sp,2
	call input
	pop word ptr y
	
	;执行函数fun
	sub sp,2
	push word ptr x
	push word ptr y
	call fun
	pop word ptr z
	
	;输出z
	printstr str3
	push word ptr z
	call output
	
	mov ax,4c00h
	int 21h

fun proc		;实现x*y+x-y
push bp
mov bp,sp
push ax
mov ax,word ptr[bp+4]
imul word ptr[bp+6]
add ax,word ptr[bp+6]
sub ax,word ptr[bp+4]
mov word ptr[bp+8],ax
pop ax
pop bp
ret 4
fun endp
input proc	;从键盘输入一个数，(-32768~32767)
push bp
mov bp,sp
push ax
push bx
push cx
push si
mov bx,10
mov word ptr[bp+4],0
mov si,1
again:mov ah,01h
int 21h
cmp al,45
jnz f1
mov si,-1
jmp again
f1:cmp al,0dh
je next
xor cx,cx
add cl,al
sub cl,30h
mov ax,word ptr [bp+4]
mul bx
add ax,cx
mov word ptr[bp+4],ax
jmp again
next:mov ax,word ptr[bp+4]
imul si
mov word ptr[bp+4],ax
pop si
pop cx
pop bx
pop ax
pop bp
ret
input endp

output proc	;将数输出到屏幕(-32768~32767)
push bp
mov bp,sp
push ax
push bx
push dx
mov bx,-1
push bx
mov bx,10
mov ax,word ptr [bp+4]
add ax,0
jns l1
mov dl,45
mov ah,02h
int 21h
mov ax,word ptr [bp+4]
neg ax
l1: cmp ax,0
jz l2
xor dx,dx
div bx
add dl,30h
push dx
jmp l1
l2:pop dx
cmp dx,-1
jz done
mov ah,02h
int 21h
jmp l2
done:pop dx
pop bx
pop ax
pop bp
ret 2
output endp

code ends
end start

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assume cs:code,ds:data,ss:stack
stack segment stack
db 1024 dup(?)
stack ends
data segment
str1 db 'please input x: $'
str2 db 'please input y: $'
str3 db 'z = $'
x dw ?
y dw ?
z dw ?
data ends
code segment
newline macro	;宏--用于输出回车换行
push ax
push dx
mov dl,0dh
mov ah,02h
int 21h
mov dl,0ah
mov ah,02h
int 21h
pop dx
pop ax
endm
printstr macro str	;宏--用于输出字符串
push ax
push dx
lea dx,str
mov ah,09h
int 21h
pop dx
pop ax
endm
start:
	mov ax,data
	mov ds,ax
	mov ax,stack
	mov ss,ax
	;用来输入x,y
	printstr str1
	sub sp,2
	call input
	pop word ptr x
	printstr str2
	sub sp,2
	call input
	pop word ptr y
	
	;执行函数fun
	sub sp,2
	push word ptr x
	push word ptr y
	call fun
	pop word ptr z
	
	;输出z
	printstr str3
	push word ptr z
	call output
	
	mov ax,4c00h
	int 21h

fun proc		;实现x*y+x-y
push bp
mov bp,sp
push ax
mov ax,word ptr[bp+4]
imul word ptr[bp+6]
add ax,word ptr[bp+6]
sub ax,word ptr[bp+4]
mov word ptr[bp+8],ax
pop ax
pop bp
ret 4
fun endp
input proc	;从键盘输入一个数，(-32768~32767)
push bp
mov bp,sp
push ax
push bx
push cx
push si
mov bx,10
mov word ptr[bp+4],0
mov si,1
again:mov ah,01h
int 21h
cmp al,45
jnz f1
mov si,-1
jmp again
f1:cmp al,0dh
je next
xor cx,cx
add cl,al
sub cl,30h
mov ax,word ptr [bp+4]
mul bx
add ax,cx
mov word ptr[bp+4],ax
jmp again
next:mov ax,word ptr[bp+4]
imul si
mov word ptr[bp+4],ax
pop si
pop cx
pop bx
pop ax
pop bp
ret
input endp

output proc	;将数输出到屏幕(-32768~32767)
push bp
mov bp,sp
push ax
push bx
push dx
mov bx,-1
push bx
mov bx,10
mov ax,word ptr [bp+4]
add ax,0
jns l1
mov dl,45
mov ah,02h
int 21h
mov ax,word ptr [bp+4]
neg ax
l1: cmp ax,0
jz l2
xor dx,dx
div bx
add dl,30h
push dx
jmp l1
l2:pop dx
cmp dx,-1
jz done
mov ah,02h
int 21h
jmp l2
done:pop dx
pop bx
pop ax
pop bp
ret 2
output endp


code ends
end start