在组装中刻薄无法按预期工作
Bitstuffing in assembly not working as intended
本文关键字:工作 更新时间:2023-10-16
我当前正在尝试学习汇编(Intel X86),并且我制作了一个程序,可以模拟32位单词上的位填充 ->连续每5个相同的位(5 0或5 1),插入了相反的位。为了将单词保持在其原始的32位尺寸,如果添加了馅料,则截断了较小的位。
这里有几个示例:
0000 1111 0000 1111 0000 1111 0000 1111 -> 0000 1111 0000 1111 0000 1111 0000 1111
0000 1111 0000 1111 0000 1111 0000 0000 -> 0000 1111 0000 1111 0000 1111 0000 0100
0000 1111 0000 1111 0000 0000 0000 0000 -> 0000 1111 0000 1111 0000 0100 0001 0000
因此,这是我的C 程序,它可以测试一切正常,但最后两个程序不起作用,我不知道为什么。我通过遵循该程序与IDE调试器进行的每个步骤进行了几次运行,并且它似乎完全按照我想要的操作,但是结果并不遵循...
#include <iostream>
using namespace std;
extern "C" {unsigned int bitstuffing(unsigned int a);}
int main () {
unsigned int in = 0xFFFFFFFF;
unsigned int verif = 0xFBEFBEFB;
unsigned int out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0x00000000;
verif = 0x04104104;
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0xF0F0F0F; // 0000 1111 0000 1111 0000 1111 0000 1111
verif = 0xF0F0F0F; // 0000 1111 0000 1111 0000 1111 0000 1111
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0xF0F0F00; // 0000 1111 0000 1111 0000 1111 0000 0000
verif = 0xF0F0F04; // 0000 1111 0000 1111 0000 1111 0000 0100
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0xF0F0000; // 0000 1111 0000 1111 0000 0000 0000 0000
verif = 0xF0F0410; // 0000 1111 0000 1111 0000 0100 0001 0000
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0xAAAA0000; // 1010 1010 1010 1010 0000 0000 0000 0000
verif = 0xAAAA0820; // 1010 1010 1010 1010 0000 1000 0010 0000
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
in = 0x7878000; // 0000 0111 1000 0111 1000 0000 0000 0000
verif = 0x7C1F041; // 0000 0111 1100 0001 1111 0000 0100 0001
// out = 0000 0111 1100 0111 1101 0000 0100 0001
out = bitstuffing(in);
if (out==verif) cout<<endl<<"OK: "<<hex<<out<<dec<<endl;
else cout<<endl<<"ERROR: "<<hex<<out<<dec<<endl;
return 0;
}
这是ASM程序,这是最重要的
CPU 386
%include "io.inc"
section .text
global CMAIN
;0FFFFFFFFh - 0xFBEFBEFB ok
;000000000h - 0x04104104 ok
;0F0F0F0Fh - 0xF0F0F0F ok
;0F0F0F00h - 0xF0F0F04 ok
;0F0F0000h - 0xF0F0410 ok
;0AAAA0000h - 0xAAAA0820 DOESNT WORK
;07878000h - 0x7878000 DOESNT WORK
CMAIN:
mov ebp, esp; for correct debugging
PUSH EBP
MOV EBP, ESP
MOV EAX, 07878000h;[EBP+8] ; places message (parameter) in EAX
MOV ECX, 32
MOV BL, 0 ; counts number of "0" bits
MOV BH, 0 ; counts number of "1" bits
loop1:
ROL EAX, 1
JC carry
JNC no_carry
carry:
XOR BL, BL ; resets "0" counter to 0
INC BH ; increments "1" counter
CMP BH, 5 ; if we get 5 consecutive bits of the same sign -> bitstuffing
JE stuffing_0
DEC ECX ; Decrementing ECX for loop
JNZ loop1
JZ end
no_carry:
XOR BH, BH ; resets "1" counter to 0
INC BL ; increments "0" counter
CMP BL, 5 ; if we get 5 consecutive bits of the same sign -> bitstuffing
JE stuffing_1
DEC ECX ; Decrementing ECX for loop
JNZ loop1
JZ end
stuffing_0:
XOR EDX, EDX
XOR EBX, EBX
MOV EDX, 2 ; Putting 2 in EDX for MUL operation
MUL EDX ; Multiplying EAX by 2 is like adding a 0 at the end
XOR EDX, EDX ; Resetting EDX register
DEC ECX ; Decrementing ECX twice for loop (in order to truncate bits)
DEC ECX
CMP ECX, 0
JG loop1
JLE end
stuffing_1:
XOR EDX, EDX
XOR EBX, EBX
MOV EDX, 2 ; Putting 2 in EDX for MUL operation
MUL EDX ; Multiplying EAX by 2 is like adding a 0 at the end
ADD EAX, 1 ; Adding 1 to EAX when the last bit is the zero we added is the same is adding 1 instead of zero
XOR EDX, EDX ; Resetting EDX register
DEC ECX ; Decrementing ECX twice for loop (in order to truncate bits)
DEC ECX
CMP ECX, 0
JG loop1
JLE end
end:
LEAVE
RET
因此,当我运行此程序时,它可以很好地与以下值(它们都放在eax中)
;0FFFFFFFFh - 0xFBEFBEFB ok
;000000000h - 0x04104104 ok
;0F0F0F0Fh - 0xF0F0F0F ok
;0F0F0F00h - 0xF0F0F04 ok
;0F0F0000h - 0xF0F0410 ok
,但不适用于以下
;0AAAA0000h - 0xAAAA0820 DOESNT WORK
;07878000h - 0x7878000 DOESNT WORK
如果有人可以发现问题,那将有很大的帮助!
塞满了您乘以2
的新位,这只是一种左右移动的方法。这将丢弃最重要的位,因此您不会从原始价值的最低意义上丢弃,而是您基本上用毛绒玩具覆盖以下位。
简而
可能的解决方案(gas
语法):
.intel_syntax noprefix
.global main
main:
sub esp, 12
mov [esp + 8], ebx
xor ebx, ebx
test_loop:
mov eax, [in + 4 * ebx]
mov dword ptr [esp], eax
call bitstuffing
mov [esp + 8], eax
cmp eax, [verify + 4 * ebx]
mov dword ptr [esp], offset ok
je got_fmt
mov dword ptr [esp], offset error
got_fmt:
mov eax, [in + 4 * ebx]
mov [esp + 4], eax
call printf
inc ebx
cmp ebx, 7
jb test_loop
mov ebx, [esp + 8]
add esp, 12
xor eax, eax
ret
bitstuffing:
push ebp
mov ebp, esp
push ebx
mov cl, 32 # 32 bits to go
xor eax, eax # the output
mov edx, [ebp + 8] # the input
xor bl, bl # the run count
next_bit:
dec cl # more bits?
js done # no
shl edx, 1 # consume from the input into CF
rcl eax, 1 # copy to output from CF
test bl, bl # first bit always matches
jz match
test al, 3 # do we have 00 or 11 in the low 2 bits?
jnp reset # no, start counting again
match:
inc bl
cmp bl, 5 # did 5 bits match?
jb next_bit # no, keep going
dec cl # space for stuffed bit?
js done # no
mov ebx, eax # make a copy
and ebx, 1 # isolate LSB
xor ebx, 1 # flip it
shl eax, 1 # make space for it
or eax, ebx # stuff it
reset:
mov bl, 1 # already have length 1
jmp next_bit
done:
pop ebx
mov esp, ebp
pop ebp
ret
.data
ok: .string "OK: 0x%08x => 0x%08xn"
error: .string "ERROR: 0x%08x => 0x%08xn"
in: .int 0xFFFFFFFF, 0x00000000, 0x0F0F0F0F, 0x0F0F0F00, 0x0F0F0000, 0xAAAA0000, 0x07878000
verify: .int 0xFBEFBEFB, 0x04104104, 0x0F0F0F0F, 0x0F0F0F04, 0x0F0F0410, 0xAAAA0820, 0x07C1F041
在IDEONE.com上看到它在操作中。
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