我正在使用汇编语言编写计算器，以便在x86处理器上执行。
基本上，我的计算器要求用户输入两个数字，然后指示要对它们进行哪种操作（加法，减法，乘法和除法）。
我的计算器正确地添加，减去和相乘，但无法分割。在进行划分时，我总是得到1。
然后我完成了我的应用程序代码：
section .data
; Messages
msg1 db 10,'-Calculator-',10,0
lmsg1 equ $ - msg1
msg2 db 10,'Number 1: ',0
lmsg2 equ $ - msg2
msg3 db 'Number 2: ',0
lmsg3 equ $ - msg3
msg4 db 10,'1. Add',10,0
lmsg4 equ $ - msg4
msg5 db '2. Subtract',10,0
lmsg5 equ $ - msg5
msg6 db '3. Multiply',10,0
lmsg6 equ $ - msg6
msg7 db '4. Divide',10,0
lmsg7 equ $ - msg7
msg8 db 'Operation: ',0
lmsg8 equ $ - msg8
msg9 db 10,'Result: ',0
lmsg9 equ $ - msg9
msg10 db 10,'Invalid Option',10,0
lmsg10 equ $ - msg10
nlinea db 10,10,0
lnlinea equ $ - nlinea
section .bss
; Spaces reserved for storing the values provided by the user.
opc resb 2
num1 resb 2
num2 resb 2
result resb 2
section .text
global _start
_start:
; Print on screen the message 1
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, lmsg1
int 80h
; Print on screen the message 2
mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, lmsg2
int 80h
; We get num1 value.
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 80h
; Print on screen the message 3
mov eax, 4
mov ebx, 1
mov ecx, msg3
mov edx, lmsg3
int 80h
; We get num2 value.
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 80h
; Print on screen the message 4
mov eax, 4
mov ebx, 1
mov ecx, msg4
mov edx, lmsg4
int 80h
; Print on screen the message 5
mov eax, 4
mov ebx, 1
mov ecx, msg5
mov edx, lmsg5
int 80h
; Print on screen the message 6
mov eax, 4
mov ebx, 1
mov ecx, msg6
mov edx, lmsg6
int 80h
; Print on screen the message 7
mov eax, 4
mov ebx, 1
mov ecx, msg7
mov edx, lmsg7
int 80h
; Print on screen the message 8
mov eax, 4
mov ebx, 1
mov ecx, msg8
mov edx, lmsg8
int 80h
; We get the option selected.
mov ebx,0
mov ecx,opc
mov edx,2
mov eax,3
int 80h
mov ah, [opc] ; Move the selected option to the registry ah
sub ah, '0' ; Convert from ascii to decimal
; We compare the value entered by the user to know what operation to perform.
cmp ah, 1
je add
cmp ah, 2
je subtract
cmp ah, 3
je multiply
cmp ah, 4
je divide
; If the value entered by the user does not meet any of the above
; conditions then we show an error message and we close the program.
mov eax, 4
mov ebx, 1
mov ecx, msg10
mov edx, lmsg10
int 80h
jmp exit
add:
; We keep the numbers in the registers eax and ebx
mov eax, [num1]
mov ebx, [num2]
; Convert from ascii to decimal
sub eax, '0'
sub ebx, '0'
; Add
add eax, ebx
; Conversion from decimal to ascii
add eax, '0'
; We move the result
mov [result], eax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
subtract:
; We keep the numbers in the registers eax and ebx
mov eax, [num1]
mov ebx, [num2]
; Convert from ascii to decimal
sub eax, '0'
sub ebx, '0'
; Subtract
sub eax, ebx
; Conversion from decimal to ascii
add eax, '0'
; We move the result
mov [result], eax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
multiply:
; We store the numbers in registers ax and bx
mov ax, [num1]
mov bx, [num2]
; Convert from ascii to decimal
sub ax, '0'
sub bx, '0'
; Multiply. AL = AX x BX
mul bx
; Conversion from decimal to ascii
add al, '0'
; We move the result
mov [result], al
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
divide:
; IN THIS LABEL IS THE ERROR!
; We store the numbers in registers ax and bx
mov dx, 0
mov ax, [num1]
mov bx, [num2]
; Convert from ascii to decimall
sub ax, '0'
sub bx, '0'
; Division. AX = DX:AX / BX
div bx
; Conversion from decimal to ascii
add ax, '0'
; We move the result
mov [result], ax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
; ALWAYS PRINTS 1
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
exit:
; Print on screen two new lines
mov eax, 4
mov ebx, 1
mov ecx, nlinea
mov edx, lnlinea
int 80h
; End the program
mov eax, 1
mov ebx, 0
int 80h
必须在标记“divide”中找到错误。
为什么我总是得到1作为分裂的结果？
我希望有经验的人可以帮助我。
非常感谢你们。我的计算器终于有效。这是我的最终代码：
section .data
; Messages
msg1 db 10,'-Calculator-',10,0
lmsg1 equ $ - msg1
msg2 db 10,'Number 1: ',0
lmsg2 equ $ - msg2
msg3 db 'Number 2: ',0
lmsg3 equ $ - msg3
msg4 db 10,'1. Add',10,0
lmsg4 equ $ - msg4
msg5 db '2. Subtract',10,0
lmsg5 equ $ - msg5
msg6 db '3. Multiply',10,0
lmsg6 equ $ - msg6
msg7 db '4. Divide',10,0
lmsg7 equ $ - msg7
msg8 db 'Operation: ',0
lmsg8 equ $ - msg8
msg9 db 10,'Result: ',0
lmsg9 equ $ - msg9
msg10 db 10,'Invalid Option',10,0
lmsg10 equ $ - msg10
nlinea db 10,10,0
lnlinea equ $ - nlinea
section .bss
; Spaces reserved for storing the values provided by the user.
opc: resb 2
num1: resb 2
num2: resb 2
result: resb 2
section .text
global _start
_start:
; Print on screen the message 1
mov eax, 4
mov ebx, 1
mov ecx, msg1
mov edx, lmsg1
int 80h
; Print on screen the message 2
mov eax, 4
mov ebx, 1
mov ecx, msg2
mov edx, lmsg2
int 80h
; We get num1 value.
mov eax, 3
mov ebx, 0
mov ecx, num1
mov edx, 2
int 80h
; Print on screen the message 3
mov eax, 4
mov ebx, 1
mov ecx, msg3
mov edx, lmsg3
int 80h
; We get num2 value.
mov eax, 3
mov ebx, 0
mov ecx, num2
mov edx, 2
int 80h
; Print on screen the message 4
mov eax, 4
mov ebx, 1
mov ecx, msg4
mov edx, lmsg4
int 80h
; Print on screen the message 5
mov eax, 4
mov ebx, 1
mov ecx, msg5
mov edx, lmsg5
int 80h
; Print on screen the message 6
mov eax, 4
mov ebx, 1
mov ecx, msg6
mov edx, lmsg6
int 80h
; Print on screen the message 7
mov eax, 4
mov ebx, 1
mov ecx, msg7
mov edx, lmsg7
int 80h
; Print on screen the message 8
mov eax, 4
mov ebx, 1
mov ecx, msg8
mov edx, lmsg8
int 80h
; We get the option selected.
mov ebx,0
mov ecx,opc
mov edx,2
mov eax,3
int 80h
mov ah, [opc] ; Move the selected option to the registry ah
sub ah, '0' ; Convert from ascii to decimal
; We compare the value entered by the user to know what operation to perform.
cmp ah, 1
je add
cmp ah, 2
je subtract
cmp ah, 3
je multiply
cmp ah, 4
je divide
; If the value entered by the user does not meet any of the above
; conditions then we show an error message and we close the program.
mov eax, 4
mov ebx, 1
mov ecx, msg10
mov edx, lmsg10
int 80h
jmp exit
add:
; We keep the numbers in the registers al and bl
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub al, '0'
sub bl, '0'
; Add
add al, bl
; Conversion from decimal to ascii
add al, '0'
; We move the result
mov [result], al
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 2
int 80h
; We end the program
jmp exit
subtract:
; We keep the numbers in the registers al and bl
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub al, '0'
sub bl, '0'
; Subtract
sub al, bl
; Conversion from decimal to ascii
add al, '0'
; We move the result
mov [result], al
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
multiply:
; We store the numbers in registers al and bl
mov al, [num1]
mov bl, [num2]
; Convert from ascii to decimal
sub al, '0'
sub bl, '0'
; Multiply. AX = AL x BL
mul bl
; Conversion from decimal to ascii
add ax, '0'
; We move the result
mov [result], ax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
divide:
; We store the numbers in registers ax and bx
mov al, [num1]
mov bl, [num2]
mov dx, 0
mov ah, 0
; Convert from ascii to decimall
sub al, '0'
sub bl, '0'
; Division. AL = AX / BX
div bl
; Conversion from decimal to ascii
add ax, '0'
; We move the result
mov [result], ax
; Print on screen the message 9
mov eax, 4
mov ebx, 1
mov ecx, msg9
mov edx, lmsg9
int 80h
; Print on screen the result
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, 1
int 80h
; We end the program
jmp exit
exit:
; Print on screen two new lines
mov eax, 4
mov ebx, 1
mov ecx, nlinea
mov edx, lnlinea
int 80h
; End the program
mov eax, 1
mov ebx, 0
int 80h
答案 0 :(得分：2)
您正在为您的号码输入读取两个字节（字符）到num1和num2。这通常是您输入的单个数字（0-9）和换行符。当你去做一个操作时，你每个读取2个字节为ax和bx，所以如果num1为5且num2为1，则ax为0xa35，bx为0xa31。然后从每个中减去0x30并除以在所有情况下给出1，然后将其转换为0x31 '1'并打印。
现在在其他情况下（add / sub），您实际上将4个字节加载到eax和ebx中。因此，当您添加5和1时，您将获得eax中的0xa310a35和ebx中的0x ???? 0a31（????来自result中的任何内容。）然而，在从每个中减去0x30并添加之后，eax的最低字节将为0x06，因此当您忽略高位字节中的内容时，您将打印6。