我在x86 Linux上使用Assembly，带有Intel 8086的说明。我的程序有问题，应该计算Fibbonacci序列的元素。
这个程序我用args运行，例如：
./ fibb 1 2 3 ，
意味着什么：序列的第一个元素是1，第二个元素是2，我们想获得第三个元素。 这个例子效果很好，但是当我尝试运行时：
./ fibb 1 2 4 ，
然后我有一些垃圾。 请帮我。我是大会的新人，所以请清楚地向我解释我做错了什么。 这是我的代码：
.intel_syntax noprefix
.global _start
.data
var1:
.ascii "To few args\n"
.equ len1, $-var1
var2:
.ascii "Wrong data\n"
.equ len2, $-var2
var3:
.ascii "wrong element of the sequence\n"
.equ len3, $-var3
var4:
.ascii "element of the sequence is higher than 255\n"
.equ len4, $-var4
var5:
.ascii "result: "
.equ len5, $-var5
var6:
.byte 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
var7:
.byte 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
var8:
.byte 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
.text
_start:
jmp ety5
ety1:
mov bh, 48
ety2:
cmp [eax], bh
jz ety3
inc bh
cmp bh, 58
jnz ety2
ety3:
cmp bh, 58
jnz ety4
inc bl
ety4:
inc eax
cmp [eax], byte ptr 0
jnz ety1
ret
ety5:
jmp ety10
invert:
xor ecx, ecx
ety6:
inc ebx
inc ecx
cmp [ebx], byte ptr 0
jnz ety6
ety7:
dec ecx
cmp ecx, 0
jz ety8
mov [ebx], byte ptr 0
dec ebx
cmp [ebx], byte ptr 48
jz ety7
inc ebx
ety8:
inc ecx
push ecx
xor ecx, ecx
ety9:
add ecx, 2
push ecx
dec ebx
mov cl, [eax]
mov ch, [ebx]
mov [eax], ch
mov [ebx], cl
inc eax
pop ecx
cmp ecx, [ebp-8]
jb ety9
pop ecx
ret
ety10:
jmp ety18
ety11:
mov cl, [eax]
cmp cl, 0
jz ety12
sub cl, 48
ety12:
mov ch, [ebx]
sub ch, 48
add cl, dh
xor dh, dh
add cl, ch
cmp cl, 10
jb ety13
inc dh
sub cl, 10
ety13:
add cl, 48
mov [eax], cl
inc eax
inc ebx
cmp [ebx], byte ptr 0
jnz ety11
ety14:
cmp [eax], byte ptr 0
jz ety16
mov cl, [eax]
sub cl, 48
add cl, dh
xor dh, dh
cmp cl, 10
jb ety15
sub cl, 10
inc dh
ety15:
add cl, 48
mov [eax], cl
inc eax
jmp ety14
ety16:
cmp dh, 0
jz ety17
add dh, 48
mov [eax], dh
ety17:
ret
ety18:
jmp ety22
ety19:
mov esi, 3
mov dl, [ecx]
sub dl, 48
loop_mul:
add dl, dl
dec esi
cmp esi,0
jnz loop_mul
add dl, dh
xor dh, dh
cmp dl, 10
jb ety20
sub dl,10
inc dh
ety20:
add dl, 48
mov [ebx], dl
inc ebx
inc ecx
cmp [ecx], byte ptr 0
jnz ety19
cmp dh, 0
jz ety21
add dh, 48
mov [ebx], dh
ety21:
mov ebx, offset var8
ret
ety22:
mov ebp, esp
mov eax, [ebp+8]
cmp eax, 0
jnz ety23
mov eax, 4
mov ebx, 1
mov ecx, offset var1
mov edx, offset len1
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety23:
mov eax, [ebp+12]
cmp eax, 0
jnz ety24
mov eax, 4
mov ebx, 1
mov ecx, offset var1
mov edx, offset len1
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety24:
mov eax, [ebp+16]
cmp eax, 0
jnz ety25
mov eax, 4
mov ebx, 1
mov ecx, offset var1
mov edx, offset len1
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety25:
xor bl, bl
mov eax, [ebp+8]
call ety1
mov eax, [ebp+12]
call ety1
cmp bl, 0
jz ety26
mov eax, 4
mov ebx, 1
mov ecx, offset var2
mov edx, offset len2
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety26:
mov eax, [ebp+16]
call ety1
cmp bl, 0
jz ety27
mov eax, 4
mov ebx, 1
mov ecx, offset var3
mov edx, offset len3
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety27:
xor ebx, ebx;
mov eax, [ebp+16]
mov ecx, [ebp+16]
ety28:
cmp [eax], byte ptr 0
jnz ety29
mov eax, 4
mov ebx, 1
mov ecx, offset var3
mov edx, offset len3
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety29:
cmp [eax], byte ptr 48
jnz ety30
inc eax
inc ecx
jmp ety28
ety30:
inc ecx
cmp [ecx], byte ptr 0
jz ety35
inc ecx
cmp [ecx], byte ptr 0
jz ety33
inc ecx
cmp [ecx], byte ptr 0
jz ety31
mov eax, 4
mov ebx, 1
mov ecx, offset var4
mov edx, offset len4
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety31:
mov cl, [eax]
inc eax
sub cl, 48
ety32:
add ebx, 100
dec cl
cmp cl, 0
jnz ety32
ety33:
mov cl, [eax]
inc eax
sub cl, 48
cmp cl, 0
jz ety35
ety34:
add ebx, 10
dec cl
cmp cl, 0
jnz ety34
ety35:
mov cl, [eax]
sub cl, 48
cmp cl, 0
jz ety37
ety36:
inc ebx
dec cl
cmp cl, 0
jnz ety36
ety37:
cmp ebx, 256
jb ety38
mov eax, 4
mov ebx, 1
mov ecx, offset var4
mov edx, offset len4
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety38:
xor dl, dl
ety39:
dec ebx
inc dl
cmp ebx,0
jnz ety39
mov eax, [ebp+12]
mov ebx, offset var7
xor dh, dh
ety40:
inc eax
inc dh
cmp [eax], byte ptr 0
jnz ety40
xor ch, ch
ety41:
dec eax
dec dh
mov cl, [eax]
mov [ebx], cl
inc ebx
cmp dh, 0
jnz ety41
mov eax, [ebp+8]
mov ebx, offset var6
ety42:
inc eax
inc dh
cmp [eax], byte ptr 0
jnz ety42
ety43:
dec eax
dec dh
mov cl, [eax]
mov [ebx], cl
inc ebx
cmp dh, 0
jnz ety43
cmp dl, 1
jnz ety44
jmp ety46
ety44:
dec dl
cmp dl, 1
jnz ety45
jmp ety47
ety45:
dec dl
cmp dl, 0
jz ety47
push edx
xor dh, dh
mov eax, offset var6
mov ebx, offset var8
mov ecx, offset var7
call ety19
xor dh, dh
call ety11
pop edx
dec dl
cmp dl, 0
jz ety46
push edx
xor dh, dh
mov eax, offset var7
mov ebx, offset var8
mov ecx, offset var6
call ety19
xor dh, dh
call ety11
pop edx
jmp ety45
ety46:
mov eax, 4
mov ebx, 1
mov ecx, offset var5
mov edx, offset len5
int 0x80
mov eax, offset var6
mov ebx, offset var6
call invert
mov eax, 4
mov ebx, 1
mov edx, ecx
mov ecx, offset var6
int 0x80
mov eax, 4
mov ebx, 1
mov [ecx], byte ptr 10
mov edx, 1
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
ety47:
mov eax, 4
mov ebx, 1
mov ecx, offset var5
mov edx, offset len5
int 0x80
mov eax, offset var7
mov ebx, offset var7
call invert
mov eax, 4
mov ebx, 1
mov edx, ecx
mov ecx, offset var7
int 0x80
mov eax, 4
mov ebx, 1
mov [ecx], byte ptr 10
mov edx, 1
int 0x80
mov eax, 1
mov ebx, 0
int 0x80
int 0x80
答案 0 :(得分：0)
这段代码应该用第一个参数（一个数字，而不是地址）填充EAX。
mov eax, [ebp+8]
call ety1
但是在 ety1 ，此代码将地址EAX 处的字节与BH进行比较。
ety1:
mov bh, 48
ety2:
cmp [eax], bh
jz ety3
这是怎么回事？
修改
当我通过一个可笑的长级跳跃到达该程序的真正入口点时，我完全忘记了这实际上是一个应用程序，而不仅仅是一些子程序。因此，参数是地址，使我的答案无效！
EDIT2
您确定DS寄存器已正确初始化吗？