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我有一個NASM代碼,它讀取一個文件(存儲在地址變量中的文件名)並計算CRC5。它佔用一個文件的每個字節並通過計算例程運行它。我觀察到一個奇怪的行爲:
如果我設置斷點後mov [curr], ebx
每第2和第3迭代curr變量設置爲0,所有其他迭代產生正確的字符。無論我打開的文本文件如何,都會發生這種情況。nasm文件緩衝區的第二和第三個元素始終爲0
SECTION .data
table dd 0x80, 0x40, 0x20, 0x10, 0x8, 0x4, 0x2, 0x1
address dd "test.cpp", 0
crc dd 0,0,0,0,0,10
size dw 8192
section .bss
doinvert: resb 1
buf resb 8192
curr resb 1
SECTION .text
global main
main:
mov ebx, address
mov eax, 5 ; open(
mov ecx, 0 ; read-only mode
int 80h ;);
mov ebx, eax ; file_descriptor,
mov eax, 3 ; read(
mov ecx, buf ; *buf,
mov edx, size ; *bufsize
int 80h ;);
mov [size], eax
mov ecx, [size]
loop_outer:
mov eax, [size]
sub eax, ecx
mov ebx, [buf+eax]
and ebx, 0ffh ; filter out extra bytes
mov [curr], ebx
push ecx
mov ecx, 8
jmp loop1
near_jump:
jmp loop_outer
loop1:
mov eax, 8
sub eax, ecx
mov ebx, [table+eax*4]
mov eax, [curr]
and ebx, eax
cmp ebx, 0
je skip
mov ebx, 1
skip:
mov eax, [crc+4*4]
xor ebx, eax
mov [doinvert], ebx
mov ebx, [crc+3*4]
mov [crc+4*4], ebx
mov ebx, [crc+2*4]
mov eax, [doinvert]
xor ebx, eax
mov [crc+3*4], ebx
mov ebx, [crc+1*4]
mov [crc+2*4], ebx
mov ebx, [crc]
mov [crc+1*4], ebx
mov ebx, [doinvert]
mov [crc], ebx
loop loop1
pop ecx
loop near_jump
mov ebx,0
mov eax,1
int 0x80
更奇怪的是,如果我將上面的代碼減少到下面的代碼字符正確迭代。
SECTION .data
table dd 0x80, 0x40, 0x20, 0x10, 0x8, 0x4, 0x2, 0x1
address dd "test.cpp", 0
crc dd 0,0,0,0,0,10
size dw 8192
section .bss
doinvert: resb 1
buf resb 8192
curr resb 1
SECTION .text
global main
main:
mov ebx, address
mov eax, 5 ; open(
mov ecx, 0 ; read-only mode
int 80h ;);
mov ebx, eax ; file_descriptor,
mov eax, 3 ; read(
mov ecx, buf ; *buf,
mov edx, size ; *bufsize
int 80h ;);
mov [size], eax
mov ecx, [size]
loop_outer:
mov eax, [size]
sub eax, ecx
mov ebx, [buf+eax]
and ebx, 0ffh
mov [curr], ebx
loop loop_outer
mov ebx,0
mov eax,1
int 0x80