我一直在尝试使用NASM和汇编语言,因此,我对子程序如何在这种语言中工作有一个基本的想法。然而,我的问题是,如何在同一程序中捆绑不同的标签。例如,我在S.O。中找到的以下代码:
;-----------------------------------------------
;SECTION .DATA
;Instantiated variables/Constants
;-----------------------------------------------
section .data
result: db "The smallest number is: " , 0x0a
result_len: equ $-result
nl: db " ", 0x0a
nl_len equ $-nl
matrix: dw 25, 24, 23, 22, 21
dw 20, 19, 18, 17, 16
dw 15, 14, 13, 12, 11
dw 10, 9, 8, 7, 6
dw 5, 4, 3, 2, 1
;-----------------------------------------------
;SECTION .BSS
;Non initialized variables
;-----------------------------------------------
section .bss
;-----------------------------------------------
;SECTION .TEXT
;Code
;-----------------------------------------------
section .text
global _start
_start:
mov edi, 0
mov esi, 0
mov ecx, 12
outerLoop:
cmp edi, 50 ;each element is 2 bytes (2 ascii characters)
ja endloop ;we need 50 because it's 5 elements per row
mov esi, 0 ;and 5 rows
innerLoop:
cmp esi, 5 ;Compare esi(inner loop index) to 5
jae innerEnd ;jump if it reached the end of the row
mov ax, [matrix + edi + esi*2]
cmp ax, cx
jg biggerThan
mov cx, ax
biggerThan:
inc esi
jmp innerLoop
innerEnd:
add edi, 10 ;row has been complete, go to next
jmp outerLoop
endloop:
push ecx
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, result_len
int 0x80
mov eax, 4
mov ebx, 1
mov ecx, esp
add [ecx], DWORD 30h
mov edx, 2
int 0x80
; display new line
mov eax, 4
mov ebx, 1
mov ecx, nl
mov edx, nl_len
int 0x80
exit:
mov eax, 1 ;eax contains 1 so quit
mov ebx, 0
int 0x80
是否可以在单个过程中捆绑innerLoop,outerLoop,greaterThan,innerEnd等,并像这样调用该过程:
;-----------------------------------------------
;SECTION .DATA
;Instantiated variables/Constants
;-----------------------------------------------
section .data
result: db "The smallest number is: " , 0x0a
result_len: equ $-result
nl: db " ", 0x0a
nl_len equ $-nl
matrix: dw 25, 24, 23, 22, 21
dw 20, 19, 18, 17, 16
dw 15, 14, 13, 12, 11
dw 10, 9, 8, 7, 6
dw 5, 4, 3, 2, 1
;-----------------------------------------------
;SECTION .BSS
;Non initialized variables
;-----------------------------------------------
section .bss
;-----------------------------------------------
;SECTION .TEXT
;Code
;-----------------------------------------------
section .text
global _start
_start:
mov edi, 0
mov esi, 0
mov ecx, 12
call findSmallestNumber
findSmallestNumber:
outerLoop:
cmp edi, 50 ;each element is 2 bytes (2 ascii characters)
ja endloop ;we need 50 because it's 5 elements per row
mov esi, 0 ;and 5 rows
innerLoop:
cmp esi, 5 ;Compare esi(inner loop index) to 5
jae innerEnd ;jump if it reached the end of the row
mov ax, [matrix + edi + esi*2]
cmp ax, cx
jg biggerThan
mov cx, ax
biggerThan:
inc esi
jmp innerLoop
innerEnd:
add edi, 10 ;row has been complete, go to next
jmp outerLoop
endloop:
push ecx
ret
mov eax, 4
mov ebx, 1
mov ecx, result
mov edx, result_len
int 0x80
mov eax, 4
mov ebx, 1
mov ecx, esp
add [ecx], DWORD 30h
mov edx, 2
int 0x80
; display new line
mov eax, 4
mov ebx, 1
mov ecx, nl
mov edx, nl_len
int 0x80
exit:
mov eax, 1 ;eax contains 1 so quit
mov ebx, 0
int 0x80
这样做会导致Linux中的分段错误错误,因此我尝试这样做的方式肯定有问题。 任何帮助将不胜感激!
答案 0 :(得分:1)
那里至少有两个问题:
push ecx
不应该是子程序的一部分,因为它已经是主程序中的代码。它通过write
系统调用设置打印缓冲区。它的位置由mov ecx, esp
行使用。请注意,ret
会从堆栈中弹出一个地址,然后返回该地址。因此,在这种情况下,它将弹出您的ecx
值,并尝试将其用作将导致错误的返回地址。call
指令后继续执行代码。在你的情况下,这意味着它将再次进入findSmallestNumber
子程序(因为它直接在call
之后),但这次在堆栈上将没有返回地址。因此,即使第1点被修复,ret
仍然会出错。解决方案是将子例程移出直接执行路径。在退出调用完成后,将整个子例程放在代码的末尾。PS:如果您打算使用汇编代码,您应该学会使用调试器,这样您就可以单步执行代码并查看问题所在。您可能还想阅读通常正在使用的calling conventions。如果您想要与其他代码连接,则需要这个。