2015-04-05 69 views
1

我对我的代码打印,我没能解决问题..打印输出错误与assymble代码

我的程序很简单敏锐地这是对如何绘制简单的线性方程实践

x在-2和2之间为常数

并且用户必须在线性等式y = ax + b中输入一个& b。

一个& b的签约用户还必须输入符号

,程序必须给5对(Y,X),但它不是为我工作

这里是代码:

{ 

    ; multi-segment executable file template. 

data segment 
    ; add your data here! 

    x db -2,-1,0,1,2 
    r1 dw 0,0,0,0,0 
    r2 dw 0,0,0,0,0 
    z1 dw 0,0,0,0,0 
    z2 dw 0,0,0,0,0 
    a1 db 0 
    a2 db 0 
    b1 db 0 
    b2 db 0 

    m1 db "This program will give u 5 pairs of result for the following eqn 'y=ax+b' where x is between -2 & 2.",0ah,0dh,"$",0ah,0dh 
    m2 db 0ah,0dh,"Enter the sign of 'a',(+,-)",0ah,0dh,"$",0ah,0dh 
    m3 db 0ah,0dh,"Enter 'a'",0ah,0dh,"$",0ah,0dh 
    m4 db 0ah,0dh,"Enter the sign of 'b',(+,-)",0ah,0dh,"$",0ah,0dh 
    m5 db 0ah,0dh,"Enter 'b'",0ah,0dh,"$",0ah,0dh 


ends 

stack segment 
    dw 128 dup(0) 
ends 

code segment 
start: 
; set segment registers: 
    mov ax, data 
    mov ds, ax 
    mov es, ax 

    ; add your code here 

    lea dx, m1 
    mov ah, 9 
    int 21h 

    lea dx, m2 
    mov ah, 9 
    int 21h 

    xor ax,ax 

    mov ah, 1  ;entering the sign of "a" 
    int 21h   

    CMP al,'-' 

    jnz next 

    lea dx, m3 
    mov ah, 9 
    int 21h 

    mov ah, 1  ;entering "a" 
    int 21h 
    sub al,30h 

    neg al 

    mov a1,al 
    xor ah,ah 
    xor si,si 
    mov cx,5 

mul1: 

    mov al,a1 

    imul x[si] 

    mov r1[si],ax ;saving the result from multiplying "-ax" 

inc si  

    loop mul1 
    jmp b 


next: ;if "a" was pasitive 

    lea dx, m3 
    mov ah, 9 
    int 21h 

    mov ah, 1  ;entering "a" 
    int 21h 
    sub al,30h 

    mov a2,al 

    xor ah,ah 
    xor si,si 
    mov cx,5 


mul2: 

    mov al,a2 

    imul x[si] 

    mov r2[si],ax ;saving the result from multiplying "ax" 

    inc si  

    loop mul2 


b: lea dx, m4 
    mov ah, 9 
    int 21h 

    mov ah, 1  ;entering the sign of "b" 
    int 21h 

    CMP al,'-' 

    jnz next1 

    lea dx, m5 
    mov ah, 9 
    int 21h 

    mov ah, 1  ;entering "b" 
    int 21h 
    sub al,30h 

    mov b1,al 

    xor ah,ah 
    xor si,si 
    mov cx,5 

sub1: 

    mov al,b1 

    sub ax,r1[si] 

    mov z1[si],ax ;saving the result from the op "ax-b" 

    inc si  

    loop sub1 

    jmp print1  

next1: 

    lea dx, m5 
    mov ah, 9 
    int 21h 

    mov ah, 1  ;entering "b" 
    int 21h 
    sub al,30h 

    mov b2,al 

    xor ah,ah 
    xor si,si 
    mov cx,5 

add1: 

    mov al,b2 

    add ax,r2[si] 

    mov z2[si],ax ;saving the result from the op "ax+b" 

    inc si  

    loop add1 

    jmp print2 

;printing the 5 pairs (x,y) when ax-b 

print1: 

    xor si,si 
    mov cx,5 

    print: 

    mov dl,0ah 
    mov ah,2 
    int 21h 

    mov dl,0dh 
    mov ah,2 
    int 21h 

    mov dl,'(' 
    mov ah,2 
    int 21h 


    lea dx,x[si] 
    mov ah,9 
    int 21h 

    mov dl,"," 
    mov ah,2 
    int 21h 

    lea dx,z1[si] 
    mov ah,9 
    int 21h 

    mov dl,")" 
    mov ah,2 
    int 21h 

    mov dl,0ah 
    mov ah,2 
    int 21h 

    mov dl,0dh 
    mov ah,2 
    int 21h 

    inc si 

    loop print 
    jmp end1 


print2: ;printing the 5 pairs (x,y) when ax+b 

    xor si,si 
    mov cx,5 

    print11: 

    mov dl,0ah 
    mov ah,2 
    int 21h 

    mov dl,0dh 
    mov ah,2 
    int 21h 

    mov dl,"(" 
    mov ah,2 
    int 21h 


    lea dx,x[si] 
    mov ah,9 
    int 21h 

    mov dl,"," 
    mov ah,2 
    int 21h 

    lea dx,z2[si] 
    mov ah,9 
    int 21h 

    mov dl,")" 
    mov ah,2 
    int 21h 

    mov dl,0ah 
    mov ah,2 
    int 21h 

    mov dl,0dh 
    mov ah,2 
    int 21h 

    inc si 

    loop print11 

end1 : 

    mov ax, 4c00h ; exit to operating system. 
    int 21h  
ends 

end start ; set entry point and stop the assembler. 

} 

和感谢提前的帮助:)

回答

0
m1 db "This program will give u 5 pairs of result for the following eqn 'y=ax+b' where x is between -2 & 2.",0ah,0dh,"$",0ah,0dh 
m2 db 0ah,0dh,"Enter the sign of 'a',(+,-)",0ah,0dh,"$",0ah,0dh 
m3 db 0ah,0dh,"Enter 'a'",0ah,0dh,"$",0ah,0dh 
m4 db 0ah,0dh,"Enter the sign of 'b',(+,-)",0ah,0dh,"$",0ah,0dh 
m5 db 0ah,0dh,"Enter 'b'",0ah,0dh,"$",0ah,0dh 

在所有这些行中,最后一个CRLF对将不会被打印,因为它在终止$之后。这是你的PRINT问题吗?

m3 db 0ah,0dh,"Enter 'a'",0ah,0dh,0ah,0dh,"$" 

这是一个问题。您将x定义为字节,将r1定义为字。您无法使用inc si以正确推进两者。我建议你定义x作为单词并使用add si,2

mul1: 
mov al,a1 
imul x[si] 
mov r1[si],ax ;saving the result from multiplying "-ax" 
inc si  
loop mul1 
+0

没有消息能够被正确打印.. – 2015-04-05 18:25:53

+0

实际的问题是在最后结果时,我想打印Ÿ – 2015-04-05 18:26:50

+0

见编辑答案。同样的问题4次。 – 2015-04-05 18:29:52