与 C++ 链接的汇编语言中的 XOR

Question

该问题与用伽马序列加密字符串的函数有关。

函数调用：

cout << "\ncipher with code: ";
    for (register int i = 0; i < str.length(); i++) {
        str[i] = fun_cipher_gamma(str[i], gamma[i]);
        cout << str[i];
    }
cout << endl;

函数本身：

char fun_cipher_gamma(register char simbol, register char gamma)
{
    char result = ' ';
    funct();
    return result;
}

汇编代码：

.MODEL FLAT, C
.STACK 256
.DATA

.CODE

    EXTRN gamma : BYTE
    EXTRN simbol : BYTE
    EXTRN result : BYTE

PUBLIC funct
funct PROC far

    mov al, simbol        
    mov bl, gamma         
    xor al, bl            
    mov result, al        
    retn

funct ENDP
END

使用调试器我意识到汇编语言函数不返回结果，但我不明白如何修复它。
我认为我需要使用push，但我不确定并且我迷失了，因为我的汇编语言技能几乎不存在

这个函数在 C++ 中运行成功，我需要用汇编语言重做它的逻辑：

return btowc(simbol) ^ btowc(gamma);

UPD：在 .cpp 中，我添加了这些外部函数：

extern "C" char funct(void);
extern "C" unsigned char gamma = ' ';
extern "C" unsigned char simbol = ' ';
extern "C" unsigned char result = ' ';

Answer 1

正如 @Chris Dodd 提到的，您没有在程序集中使用全局变量！
最好使用调用约定，而不是使用全局变量作为参数。它提高了可读性并使其更易于理解。

extern "C" char funct(unsigned char symbol, unsigned char gamma);
char fun_cipher_gamma(unsigned char symbol, unsigned char gamma)
{
    return funct(symbol, gamma);
}

; in C calling convention:
; Caller passes Parameters via stack and all arguments are pushed to stack as dwords (4 byte)
; Callee should preserve EBX, ESI, EDI, EBP, and ESP
; and use EAX to return a value
push ebp     ; Callee should preserver old value of EBP
mov ebp, esp ; set the base pointer to stack pointer
; sub esp, 8   ; make room for 8 byte of variables (commented out because we aren't using it)

; push any other registers that you'll use HERE

; ...
; [ebp - 12] = local variable #3
; [ebp -  8] = local variable #2
; [ebp -  4] = local variable #1
; [ebp +  0] = old EBP the we pushed
; [ebp +  4] = return address
; [ebp +  8] = parameter #1
; [ebp + 12] = parameter #2
; [ebp + 16] = parameter #3
; [ebp + 20] = parameter #4
; ...

; so this means "symbol" is at [ebp + 8] and gamma is at [ebp + 12]
; [ebp +  8] = symbol
; [ebp + 12] = gamma

mov eax, [ebp +  8]
xor eax, [ebp + 12]

; pop saved registers HERE

mov esp, ebp ; restore ESP to its original value
pop ebp      ; restore EBP's value
ret

但请注意，现在使用汇编不会提高程序的性能（事实上，您可能会损失性能），编译器编写的汇编比我们更快。
如果它们没有优化，则可能会编写比编译器更快的程序集，但一旦您启用优化，它们就会编写比您更快的程序集。

MSVC19 编译了这个：

; in C calling convention:
; Caller passes Parameters via stack and all arguments are pushed to stack as dwords (4 byte)
; Callee should preserve EBX, ESI, EDI, EBP, and ESP
; and use EAX to return a value
push ebp     ; Callee should preserver old value of EBP
mov ebp, esp ; set the base pointer to stack pointer
; sub esp, 8   ; make room for 8 byte of variables (commented out because we aren't using it)

; push any other registers that you'll use HERE

; ...
; [ebp - 12] = local variable #3
; [ebp -  8] = local variable #2
; [ebp -  4] = local variable #1
; [ebp +  0] = old EBP the we pushed
; [ebp +  4] = return address
; [ebp +  8] = parameter #1
; [ebp + 12] = parameter #2
; [ebp + 16] = parameter #3
; [ebp + 20] = parameter #4
; ...

; so this means "symbol" is at [ebp + 8] and gamma is at [ebp + 12]
; [ebp +  8] = symbol
; [ebp + 12] = gamma

mov eax, [ebp +  8]
xor eax, [ebp + 12]

; pop saved registers HERE

mov esp, ebp ; restore ESP to its original value
pop ebp      ; restore EBP's value
ret

对此：

xor     cl, dl
mov     al, cl
ret     0

它使用更少的空间，同时比我们的实现更快（它使用fastcall调用约定）。
有关调用约定的更多信息：https ://www.dyncall.org/docs/manual/manualse11.html