运行我的代码时,我总是收到分段错误错误。一切都编好了,但我似乎无法让它做我想做的事。该方案是要求用户输入3点的整数,然后问他们怎么想的数字的平均值将是用户,考虑到这一点,然后用该用户是否正确使用汇编代码程序发生分段错误错误
segment .data
;
; Output strings
;
prompt1 db "Enter a positive integer: ", 0
prompt2 db "Enter a second positive integer: ", 0
prompt3 db "Enter a third positive integer: ", 0
prompt4 db "Enter a guess of their average: ", 0
outmsg1 db "You entered ", 0
outmsg2 db " and ", 0
outmsg3 db " and ", 0
outmsg4 db "You guessed that the average is ", 0
outmsg5 db "You did you guess correctly? (0 = no, 1 = yes)", 0
avermsg db "The average of the numbers is ", 0
segment .bss
input1 resd 1
input2 resd 1
input3 resd 1
input4 resd 1
guess resd 1
segment .text
Global main
main:
enter 0,0 ; setup routine
pusha
mov eax, prompt1 ; print out prompt1
call print_string
call read_int ; read integer
mov [input1], eax ; store integer into input1
mov eax, prompt2 ; print out prompt2
call print_string
call read_int ; read integer
mov [input2], eax ; store integer into input2
mov eax, prompt3 ; print out prompt3
call print_string
call read_int ; read integer
mov [input3], eax ; store integer into input3
mov eax, prompt4 ; print out prompt4
call print_string
call read_int ; read integer
mov [guess], eax
mov eax, [input1] ; eax = dword at input1
add eax, [input2] ; eax += dword at input2
add eax, [input3] ; eax += dword at input3
mov ebx, 3
div ebx ; divides the sum by 3
mov ecx, eax ; freeing up eax, puts quotient into ecx
dump_regs 1 ; print out register values
; next print out results
mov eax, outmsg1
call print_string ; print out first message
mov eax, [input1]
call print_int
mov eax, outmsg2
call print_string ; print out second message
mov eax, [input2]
call print_int
mov eax, outmsg3
call print_string ; print out thrid message
mov eax, [input3]
call print_int
mov eax, outmsg4
call print_string ; print out fourth message
mov eax, [input4]
call print_int
xor ebx, ebx
cmp ecx, [guess]
sete bl
neg ebx
mov edx, ebx
and ecx, edx
not ebx
and ebx, [guess]
or edx, ebx
mov eax, outmsg5
call print_string
mov ecx, eax
call print_int
mov eax, [avermsg]
call print_string ; print out final message
mov ecx, edx
call print_int ; print out average of ebx
call print_nl ; print new line
popa
mov eax, 0 ; return back to C
leave
ret
omg为什么要装配 – dynamic 2011-03-19 20:37:35
请格式化代码,以便它在这里可读,并确定它发生故障的确切位置。 asm很难阅读,因为它是。 – Mat 2011-03-19 20:40:41
感谢digitalRoss,我不确定分段故障发生在哪里,它在运行时发生。 – BDilly 2011-03-19 20:44:18