如何修复十六进制转二进制汇编程序出现反向转换的问题？

First, let's break down what's going wrong here — your current code is built to read binary digits (0/1), pack them into a single byte, then output that byte as a hexadecimal character. No wonder it's backwards! Let's fix this step by step to make it convert hex input to binary output, plus track those 1's and 0's counts properly.

Key Modifications Needed #

1. Update Input Logic to Accept Hexadecimal Characters #

Your original input only allows 0/1. We need to expand this to accept valid hex characters: 0-9, A-F (uppercase), and a-f (lowercase). We'll also handle up to 2 hex digits since a byte is 8 bits, and 2 hex digits equal exactly 8 bits.

2. Rewrite Output to Print Binary Digits #

Instead of converting the byte to a hex character, we'll loop through each of the 8 bits in the byte, print '0' or '1' for each bit, and count how many of each we encounter.

3. Implement Proper Count Tracking #

Your code declares countone and countzero but never uses them. We'll initialize these to 0, increment them as we print each bit, then convert the counts to ASCII for readable output.

Full Fixed Code #

.model small
.stack
.data
msg1 db "Enter hexadecimal number (1-2 digits):$"
msg2 db "The binary value is:$"
msg3 db "The number of 1's is:$"
msg4 db "The number of 0's is:$"
countone db ?
countzero db ?
.code
main proc
mov ax,@data
mov ds,ax

clear:
; Print newline
mov ah,2
mov dl,0dh
int 21h
mov dl,0ah
int 21h

; Prompt for hex input
mov ah,9
lea dx,msg1
int 21h

xor bh,bh ; Initialize bh to 0 (will hold our 8-bit value)
mov cl,0 ; Track number of hex digits entered (max 2)

input:
mov ah,1
int 21h
mov ch,al

cmp ch,0dh ; Check if user pressed enter
je print

; Validate and convert hex character
cmp ch, '0'
jl exit ; Invalid character, exit
cmp ch, '9'
jle convert_digit

cmp ch, 'A'
jl check_lower
cmp ch, 'F'
jle convert_upper

check_lower:
cmp ch, 'a'
jl exit
cmp ch, 'f'
jg exit
convert_lower:
sub ch, 'a' - 10 ; Convert a-f to 10-15
jmp process_hex

convert_upper:
sub ch, 'A' - 10 ; Convert A-F to 10-15
jmp process_hex

convert_digit:
and ch, 0Fh ; Convert 0-9 to 0-9 (same as sub ch, '0')

process_hex:
shl bh, 4 ; Shift existing bits left to make space for new 4 bits
or bh, ch ; Add the new 4-bit value to bh
inc cl
cmp cl,2 ; Stop after 2 hex digits
jl input

print:
; Print newline
mov ah,2
mov dl,0dh
int 21h
mov dl,0ah
int 21h

; Print binary value prompt
mov ah,9
lea dx,msg2
int 21h

; Initialize counts to 0
mov countone, 0
mov countzero, 0

; Print 8 binary bits
mov cx,8
print_bit:
shl bh,1 ; Shift left, carry flag holds the highest bit
jc print_one ; If carry is set, it's a 1

print_zero:
mov dl,'0'
inc countzero
jmp print_char

print_one:
mov dl,'1'
inc countone

print_char:
mov ah,2
int 21h
loop print_bit

; Print newline
mov ah,2
mov dl,0dh
int 21h
mov dl,0ah
int 21h

; Print number of 1's
mov ah,9
lea dx,msg3
int 21h
mov al,countone
add al,'0' ; Convert count to ASCII
mov dl,al
mov ah,2
int 21h

; Print newline
mov ah,2
mov dl,0dh
int 21h
mov dl,0ah
int 21h

; Print number of 0's
mov ah,9
lea dx,msg4
int 21h
mov al,countzero
add al,'0'
mov dl,al
mov ah,2
int 21h

jmp clear

exit:
; Print newline before exiting
mov ah,2
mov dl,0dh
int 21h
mov dl,0ah
int 21h
mov ah,4ch
int 21h
main endp
END main

What Changed? #

• Input Handling: Now accepts 1-2 hex digits (0-9, A-F, a-f) and packs them into an 8-bit value in bh
• Binary Output: Loops through each bit of bh, prints '0'/'1', and increments the respective count
• Count Tracking: Properly initializes and updates countone/countzero, then converts the counts to ASCII for readable output
• User Feedback: Updated the input prompt to clarify we accept 1-2 hex digits

内容的提问来源于stack exchange，提问作者eljonc