汇编:字符串输入(字大小)和重新打印

时间:2012-09-19 22:29:03

标签: assembly x86 dos 16-bit

我不知道出了什么问题。我想以WORD大小输入字符串,并在输入的字符串值中添加一个整数并重新打印结果。我不太确定算术运算,因为这是我第一次在一个程序中使用很多操作,它们是16位。

clr macro  
mov ax, 03h  
int 10h  
endm  

cseg segment para 'code'  
assume cs:cseg; ds:cseg; ss:cseg; es:cseg  
org 100h  

start: jmp begin  

amount_s label word  
amount_max dw 3  
amount_length dw ?  
amount_field dw 3 dup (?)  

x1 dw 0  
x2 dw 0  

sum1 dw 0
sum2 dw 0

bal dw 10

begin:  clr

mov ah, 0Ah     ;input string
lea dx, amount_s
mov cx, amount_length
lea si, amount_field
int 21h

mov ax, [si]        ;copy si to ax
sub ax, 30h     ;converts value of ax to integer
mov bx, 10      ;copy 10 to bx      
mul bx          ;multiply it ax by bx
mov x1, ax      ; copy ax to x1
inc si          ;move si pointer by 1

mov ax, [si]        ;copy si to ax
sub ax, 30h     ;converts value of ax to integer
mov x2, ax      ; copy ax to x2

add ax, x1      ;add ax which is x2 by x1
add ax, bal     ; add ax by bal which is 10
mov sum1, ax        ;copy the result to sum1

mov dx, 0       ; copy 0 to dx  
mov bx, 10      ; copy 10 to bx
div bx          ;divides ax by bx
mov sum1, ax        ; copy quotient to sum1
mov sum2, dx        ; copy remainder to sum2

add sum1, 30h       ;convert for printing
add sum2, 30h       ;convert for printing

mov ah, 02h     ;prints sum1
mov dx, sum1
int 21h

mov ah, 02h     ;prints sum2
mov dx, sum2
int 21h

int 20h
cseg ends
end start

1 个答案:

答案 0 :(得分:2)

如何调试和检查参考文档,看看出了什么问题以及如何? :)

我可以直截了当地说,int 21h的函数0ah使用的结构包含字节字段,而不是字段。

然而你将它们声明为单词(dw),而不是字节(db):

amount_max dw 3
amount_length dw ?
amount_field dw 3 dup (?)

你不应该像以下那样访问它们:

mov ax, [si]

相反,读取字节:

mov al, [si]

如果你想将字节值转换成字值,只需将0加到字寄存器的顶部字节中,如下所示:

mov ah, 0

其余的看起来很合理,但我还没有运行代码。你应该这样做。在调试器中,如果它无法正常工作。