我正在编写一个程序来从用户那里获取一个整数,然后打印出从0到数字的所有数字。我的代码输入正常,但是当打印出来时,它会在看似无限循环的情况下连续打印。这是我的代码:
SECTION .data ; Constant variable declaration
len EQU 32 ; Constant length
msg db "Enter a number: ", 0 ; Input message
msglen EQU $-msg ; Input message length
SECTION .bss ; Uninitialised data declaration
other resd len ; Output counter that is incremented
data resd len ; Input data buffer
SECTION .text ; Main program initialiser
GLOBAL _start ; Linker entry point declaration
_start: ; Entry point
nop ; This keeps the debugger happy :)
Msg: ; This section prints out the message
mov eax, 4 ; }
mov ebx, 1 ; }
mov ecx, msg ; } System_write call
mov edx, msglen ; }
int 80h ; }
input: ; This section gets the integer from the user
mov eax, 3 ; }
mov ebx, 0 ; }
mov ecx, data ; } System_read call
mov edx, len ; }
int 80h ; }
ASCIIAdj:
mov ebp, 48 ; This line sets the counter to '0' ASCII
setup: ; This section adjusts the counter
mov [other], ebp ; Increment counter
loop: ; This section loops, printing out from zero to the number given
mov eax, 4 ; }
mov ebx, 1 ; }
mov ecx, other ; } System_write call
mov edx, len ; }
int 80h ; }
mov eax, 1 ; Move 1 to eax
add ebp, eax ; Add eax to ebp (essentially increment ebp)
mov eax, other ; move other to eax
mov ebx, data ; move data to ebx
cmp eax, ebx ; compare them
jne setup ; If they are not the same, go back to the setup to increment other
exit: ; Exits the program
mov eax, 1 ; }
mov ebx, 0 ; } System_exit call
int 80h ; }
为什么它会连续循环?我已经递增了计数器,并将输入和计数器进行了比较,为什么它不会中断?
提前致谢
编辑: 预期产出:
Enter a number: 6
0123456
该计划的一般语义:
Display "Enter a number: "
Read in an integer less than 32 bytes in size.
Set a counter variable to the ASCII value of zero
Loop:
Display the character, adding 1 to it, and checking to see if it is equal to the value inputted.
If it is equal, goto the exit section and exit
Else loop.
答案 0 :(得分:1)
这正在把我的记忆深深地挖回到我记忆深处的深处,但我想你想要
mov eax, [other] ; move other to eax
mov ebx, [data] ; move data to ebx
请注意代码中缺少的括号。您正在将other
和data
的地址加载到eax
和ebx
,而不是其中包含的值。