Question

首先发帖，请温柔。我的任务是获取二进制数，将其转换为八进制数，并显示所述八进制数。我得到了下面的代码供参考，它转换为十六进制，但我不能在我的生活中找到有关如何从二进制转换为八进制的任何文档。我的同龄人也很难过，任何帮助或煽动都会受到慷慨的赞赏。谢谢！

; program to do octal input and output
org 100h
section .data
prompt1: db "Please enter a decimal number: $"
prompt2: db 0Dh,0Ah, "The number in octal is:    $"
prompt3: db 0Dh,0Ah, "The number in decimal is:    $"

section .text
mov ah,9           ; print prompt
mov dx,prompt1
int     21h
call    dec_in      ; read value into bx
mov ah,9            ; print output label
mov dx,prompt2
int     21h 
call    hexout
mov ah,9            ; print output label
mov dx,prompt3
int     21h
call    dec_out     ; display the value in bx as hex
exit:
; exit to DOS
mov ax,4C00h        ; Normal Exit
int 21h             ; bye!

; dec_in will read a base 10 value from the keyboard and place it into the bx    register
dec_in: 
; save registers
push    ax
push    dx

xor bx,bx       ; bx holds accumulated input
mov ah,1        ; read char fcn
int 21h         ; read it into al
while1: 
cmp al,0Dh      ; char = CR?
je  finis       ; if so, we are done
push    ax      ; save the character read
mov ax,10       ; set up for multiply
mul bx          ; dx:ax <- bx * 10
mov bx,ax       ; put 16-bit result back in bx (assume no overflow)
pop ax          ; restore the char read
and ax,000Fh    ; convert character '0'-'9' to value 0-9
add bx,ax       ; add value to accumulated input
mov ah,1        ; read char fcn
int 21h         ; read next char into al
jmp while1      ; loop until done
finis:  
; restore registers
pop dx
pop ax
ret


; hexout will display the binary value in the bx register as a base 16 value    
hexout:
; save registers we will be using
push    ax
push    cx
push    dx
mov ah,2        ; display char fcn
mov cx,4        ; loop counter init
for1:               ; top of for loop
rol bx,4        ; rotate so digit is in lowest 4 bits
mov dl,bl       ; get low half in dl
and dl,0Fh      ;  and mask out all but 4 bits
cmp dl,9        ; dl <= 9?
jnbe    AtoF    ; if not, then it's A-F
or  dl,30h      ; convert 0-9 to '0'-'9'
jmp endif1      ; get ready to display
AtoF:   add dl,55   ; convert 10-15 to 'A'-'F'
endif1: int 21h     ; display char
loop    for1    ; loop until done
; restore registers
pop dx
pop cx
pop ax
ret

; dec_out will display the binary value in the bx register as a base 10 value   
dec_out:
; save registers we will be using
push    ax
push    bx
push    cx
push    dx

xor cx,cx       ; cx counts digits, initially zero
rept:
mov ax,bx       ; set up to divide by by 10
xor dx,dx       ; must have a 32 bit (unsigned) dividend
mov bx,10       ; divisor will be in bx
div bx          ; quotient will be in ax, remainder in dx
push    dx      ; push remainder on stack
inc cx          ; we generated another digit, so count it
mov bx,ax       ; the quotient goes back in bx
cmp ax,0        ; clever way to test if quotient is zero
jne rept        ; if not, generate next digit

mov ah,2        ; display character function
for2:               ; loop cx times
pop dx          ; pop digit to print
or  dl,30h      ; convert the digit to print to ASCII code
int 21h         ; display the character
loop    for2    ; and keep going until all digits displayed

; restore registers
pop dx
pop cx
pop bx
pop ax
ret

Answer 1

首先，如果您正在寻找有关数字基数转换的文档，您应该查看小学数学书籍;）您所需要的只是由基数重复划分以转换为任何基数。

然而，从二进制转换为八进制是一种更容易的特殊情况，因为8是2的幂。这里，您可以简单地将每3位转换为八进制数。

Answer 2

八进制数只是二进制数的表示，使其成为shorter and more illustrative。精确的3位转换为八进制数字。

让我们用二进制数01101010

来做

首先将三元组中的数字分组（like a decimal number）：01,101,010。第一组只有两位数，因此添加一个前导零：001,101,010。现在将其转换为十进制：1,5,2。这些十进制数字也是八进制数字，因此八进制结果为152。

如果您有“计算机表格”中的数字，您可以通过移动或类似方式直接访问二进制数字。

将二进制转换为八进制汇编Lang

2 个答案: