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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。为了使字保持其原始的 32 位大小，如果添加填充位，则将截断较低的有效位。

这里有一些例子：

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

所以这是我的 C++ 程序，它测试一切是否正常，但最后两个不起作用，我不知道为什么。我按照程序使用 IDE 调试器执行的每一步运行了它几次，它似乎完全按照我的意愿执行，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;
}

这就是 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

所以当我运行这个程序时，它可以很好地使用以下值（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

如果有人能发现问题，那将有很大帮助！

4

1 回答 1

1

填充你乘以的新位，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丢弃，而是基本上用填充位覆盖以下位。

简而言之，您的代码并没有按照您所说的那样做:)

一个可能的解决方案（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

在 ideone.com 上查看它的运行情况。

于 2015-05-11T01:48:20.570 回答