我正在编写一个子程序，只是使用堆栈将十进制数字重新打印为字符串，但是没有得到我期望的值。当我通过调试器运行时，我发现我无法从esi中获取值。我怀疑我不能以我自己的方式使用esi，但我不确定在另一种方式我能做到这一点。另外，我不允许将存储在edx中的元素推送到堆栈中。
子程序代码：
%define STDIN 0
%define STDOUT 1
%define SYSCALL_EXIT 1
%define SYSCALL_READ 3
%define SYSCALL_WRITE 4
%define BUFLEN 256
SECTION .bss ; uninitialized data section
src_str: resb BUFLEN ; buffer for backwards number
dec_str: resb BUFLEN ; number will be converted and put in this buffer
rlen: resb 4 ; length
SECTION .text ; code begins here
global prt_dec
; Begin subroutine
prt_dec:
push eax
push ebx
push ecx
push edx
push esi
push edi
mov eax, [esp + 28] ; store the decimal number 4 bytes each for each push, plus the eip
mov esi, src_str ; point esi to the backwards string buffer
mov edi, dec_str ; point edi to the new buffer
mov ebx, 10 ; stores the constant 10 in ebx
div_loop:
mov edx, 0 ; clear out edx
div ebx ; divide the number by 10
add edx, '0' ; convert from decimal to char
mov [esi], edx ; store char in output buffer
inc esi ; move to next spot in output buffer
inc ecx ; keep track of how many chars are added
cmp eax, 0 ; is there anything left to divide into?
jne div_loop ; if so, continue the loop
output_loop:
add esi, ecx ; move 1 element beyond the end of the buffer
mov al, [esi - 1] ; move the last element in the buffer into al
mov [edi], al ; move it into the first position of the converted output buffer
inc edi ; move to the next position of the converted output buffer
dec ecx ; decrement to move backwards through the output buffer
cmp ecx, 0 ; if it doesn't equal 0, continue loop
jne output_loop
print:
mov eax, SYSCALL_WRITE ; write out string
mov ebx, STDOUT
mov ecx, dec_str
mov edx, 0
mov edx, rlen
int 080h
pop_end:
pop edi ; move the saved values back into their original registers
pop esi
pop edx
pop ecx
pop ebx
pop eax
ret
; End subroutine
驱动：
%define STDIN 0
%define STDOUT 1
%define SYSCALL_EXIT 1
%define SYSCALL_READ 3
%define SYSCALL_WRITE 4
SECTION .data ; initialized data section
lf: db 10 ; just a linefeed
msg1: db " plus "
len1 equ $ - msg1
msg2: db " minus "
len2 equ $ - msg2
msg3: db " equals "
len3 equ $ - msg3
SECTION .text ; Code section.
global _start ; let loader see entry point
extern prt_dec
_start:
mov ebx, 17
mov edx, 214123
mov edi, 2223187809
mov ebp, 1555544444
push dword 24
call prt_dec
add esp, 4
call prt_lf
push dword 0xFFFFFFFF
call prt_dec
add esp, 4
call prt_lf
push 3413151
call prt_dec
add esp, 4
call prt_lf
push ebx
call prt_dec
add esp, 4
call prt_lf
push edx
call prt_dec
add esp, 4
call prt_lf
push edi
call prt_dec
add esp, 4
call prt_lf
push ebp
call prt_dec
add esp, 4
call prt_lf
push 2
call prt_dec
add esp, 4
mov eax, SYSCALL_WRITE ; write message
mov ebx, STDOUT
mov ecx, msg1
mov edx, len1
int 080h
push 3
call prt_dec
add esp, 4
mov eax, SYSCALL_WRITE ; write message
mov ebx, STDOUT
mov ecx, msg3
mov edx, len3
int 080h
push 5
call prt_dec
add esp, 4
call prt_lf
push 7
call prt_dec
add esp, 4
mov eax, SYSCALL_WRITE ; write message
mov ebx, STDOUT
mov ecx, msg2
mov edx, len2
int 080h
push 4
call prt_dec
add esp, 4
mov eax, SYSCALL_WRITE ; write message
mov ebx, STDOUT
mov ecx, msg3
mov edx, len3
int 080h
push 3
call prt_dec
add esp, 4
call prt_lf
; final exit
;
exit: mov EAX, SYSCALL_EXIT ; exit function
mov EBX, 0 ; exit code, 0=normal
int 080h ; ask kernel to take over
; A subroutine to print a LF, all registers are preserved
prt_lf:
push eax
push ebx
push ecx
push edx
mov eax, SYSCALL_WRITE ; write message
mov ebx, STDOUT
mov ecx, lf
mov edx, 1 ; LF is a single character
int 080h
pop edx
pop ecx
pop ebx
pop eax
ret
答案 0 :(得分：1)
我想到的修正（星号表示我触摸的线条），希望从评论中我可以清楚地看到：
...
div_loop:
* xor edx, edx ; clear out edx
div ebx ; divide the number by 10
* add dl, '0' ; convert from decimal to char
* mov [esi], dl ; store char in output buffer
inc esi ; move to next spot in output buffer
inc ecx ; keep track of how many chars are added
* test eax,eax ; is there anything left to divide into?
* jnz div_loop ; if so, continue the loop
* ; (jnz is same instruction as jne, but in this context I like "zero" more)
* mov [rlen], ecx ; store number of characters into variable
output_loop:
* ; esi already points beyond last digit, as product of div_loop (removed add)
* dec esi ; point to last/previous digit
mov al, [esi] ; move the char from the div_loop buffer into al
mov [edi], al ; move it into the first position of the converted output buffer
inc edi ; move to the next position of the converted output buffer
dec ecx ; decrement to move backwards through the output buffer
* jnz output_loop ; if it doesn't equal 0, continue loop
print:
mov eax, SYSCALL_WRITE ; write out string
mov ebx, STDOUT
mov ecx, dec_str
* mov edx, [rlen] ; read the number of digits from variable
int 080h
...