Question

美好的一天，

我一直在尝试编写一个执行以下操作的程序：

接受来自用户的两个ASCII号码（我还没有费心去尝试检查值）
将这些ASCII数转换为十进制值。
从这两个数字中找出最大公约数
显示结果

我觉得我已经成功完成了第1-3步，尽管我并不完全确定。这是代码：

bits 16
org 0x100;
jmp main ;

number1_str: db 3,0,0,0,0,'$'
number2_str: db 3,0,0,0,0,'$'
num1_hex: dw 2
num2_hex: dw 2

prompt1: db 'Please enter first number (from 1 to 50): ','$'
prompt2: db 'Please enter second number (from 1 to 50): ','$'

cr_lf:
     db 13,10,'$'   ; carriage return and line feed

;
;Displays a string in dx
;
disp_str: 
    mov ah,09              ;
    int 0x21               ;
    ret                ;

;
; Converts an ASCII string of digits to a decimal number, and puts the result
; in ax, and passes the address to dx. If an error occurs, AL is set to 'E'.
;
str_to_num:
    xor ax,ax              ; initial value of AX = 0
    xor bh,bh              ;                  BH = 0
    mov cx,10              ; To build integer in AX (multiply by 10)
    mov si,dx              ; DX points to start of input buffer
    call next_char         ;  
    ret                ;
next_char:
    mov bl,[si]            ; move contents of memory pointed to by SI to BL
    cmp bl,0x0D    ; Is it a carriage return?
    je finis               ; Yes, we are done
    cmp bl,0x39            ; ASCII for the character '9' is 39h
    jg  error              ; > '9', invalid character
    sub bl,0x30            ; Convert to numeric value (ASCII '0' - 30h)
    jl error               ; < 0, invalid character
    imul cx        ; DX:AX = AX * 10 (32-bit result)
    add ax,bx              ; add next digit
    inc si                 ; pointer to next char
    jmp next_char      ; repeat for next character
    ret                ;
error: 
    mov al,'E'             ; Flag an error
finis:
    ret                ; return to calling program

;
;Calculates the greatest common divisor from the values in
;ax and bx. The GCD will be stored in ax.
;
GCD:
    idiv bx         ; remainder (7) in DX, quotient (1) in ax
    mov ax,bx               ; move bx into ax
    mov bx,dx               ; move dx into bx
    cmp bx, 0x0     ; Is y = 0?
    jg yIsZero              ; Y is zero, return to call from main
    call GCD                ; loop again if y isn't zero
;
; Displays result in ax and terminates program
;
yIsZero:
    xor dx,dx              ; set dx to 0
    add ax,0x30    ; convert remainder to ascii, store in ax
    mov dx,ax              ;
    int 0x21               ;
    int 0x20               ; terminate program


main:
    mov dx,prompt1     ; move prompt1 to dx
    call disp_str              ; display prompt1

    ;get number1_str
    xor dx,dx              ; set dx to 0
    mov ah,0x0a    ; accept string from user
    mov dx,number1_str; address for string
    int 0x21               ;    

    call str_to_num    ;
    mov [num1_hex],ax ; put value of ax into contents of memory address at
                        ; num1_hex

    mov ah,09              ; display string
    mov dx,cr_lf       ; display carriage return and line feed
    int 0x21               ;

    mov dx,prompt2     ; move prompt2 to dx
    call disp_str      ; display prompt2

    ;get number2
    xor dx,dx              ; set dx to 0
    mov ah,0x0a    ; accept string from user
    mov dx,number2_str; address for string
    int 0x21               ;

    call str_to_num    ;
    mov [num2_hex],ax ; put value of ax into contents of memory address at
                        ; num2_hex

    mov ah,09              ; display string
    mov dx,cr_lf       ; display carriage return and line feed
    int 0x21               ;

    ; Find the greatest common divisor
    xor dx,dx              ; set dx to 0
    mov ax,num1_hex   ; move value in num1_hex to ax
    mov bx,num2_hex   ; move value in num2_hex to bx
    call GCD               ; find greatest common divisor from values in ax and bx

我期待y_is_zero显示结果，但它只是在我运行上面的代码时等待输入。如果我按任意键，程序就会结束。我

非常感谢任何帮助。 - JS

Answer 1

GDC似乎没有尝试显示任何内容;它从不调用disp_str，它所做的系统调用（在退出程序之前）与该子例程中的那个不匹配。

在nasm中从2个输入数字中查找并显示最大公约数

1 个答案: