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c++ - 汇编中的位填充未按预期工作

转载 作者:行者123 更新时间:2023-11-28 06:16:35 25 4
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我目前正在尝试学习汇编(Intel x86)并且我制作了一个程序来模拟 32 位字上的位填充 -> 每 5 个连续的相同位(5 个 0 或 5 个 1),插入一个相反的位。为了保持字的原始 32 位大小,如果添加填充位,则将截断较低有效位。

这里有几个例子:

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

这是我的 C++ 程序,用于测试是否一切正常,但最后两个不工作,我不明白为什么。我通过使用 IDE 调试器跟踪程序执行的每一步来多次运行它,它似乎完全按照我想要的方式执行,但结果不符合...

#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;
}

这是最重要的ASM程序

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

所以当我运行这个程序时,它可以很好地处理以下值(它们都放在 EAX 中)

;0FFFFFFFFh - 0xFBEFBEFB ok
;000000000h - 0x04104104 ok
;0F0F0F0Fh - 0xF0F0F0F ok
;0F0F0F00h - 0xF0F0F04 ok
;0F0F0000h - 0xF0F0410 ok

但不适用于以下内容

;0AAAA0000h - 0xAAAA0820 DOESNT WORK
;07878000h - 0x7878000 DOESNT WORK

如果有人能发现问题,那将有很大帮助!

最佳答案

要填充新位,您需要乘以 2,这只是左移的一种复杂方式。这将丢弃最高有效位,因此您不会从原始值的最低有效位置开始丢弃,相反,您基本上是用填充位覆盖后续位。

简而言之,您的代码并没有按照您所说的去做:)

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%08x\n"
error: .string "ERROR: 0x%08x => 0x%08x\n"
in: .int 0xFFFFFFFF, 0x00000000, 0x0F0F0F0F, 0x0F0F0F00, 0x0F0F0000, 0xAAAA0000, 0x07878000
verify: .int 0xFBEFBEFB, 0x04104104, 0x0F0F0F0F, 0x0F0F0F04, 0x0F0F0410, 0xAAAA0820, 0x07C1F041

查看in operation at ideone.com .

关于c++ - 汇编中的位填充未按预期工作,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30158181/

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