我已经创建了这个程序一段时间了，我无法弄清楚如何解决解密例程，任何帮助将不胜感激。
截至目前，代码的加密部分正常工作。
#include <conio.h> // for kbhit
#include <iostream> // for cin >> and cout <<
#include <iomanip> // for fancy output
using namespace std;
#define MAXCHARS 6 // feel free to alter this, but 6 is the minimum
#define dollarchar '$' // string terminator
char OChars[MAXCHARS],
EChars[MAXCHARS],
DChars[MAXCHARS] = "Soon!"; // Global Original, Encrypted, Decrypted character strings
//----------------------------- C++ Functions ----------------------------------------------------------
void get_char (char& a_character)
{
cin >> a_character;
while (((a_character < '0') | (a_character > 'z')) && (a_character != dollarchar))
{ cout << "Alphanumeric characters only, please try again > ";
cin >> a_character;
}
}
//-------------------------------------------------------------------------------------------------------------
void get_original_chars (int& length)
{ char next_char;
length = 0;
get_char (next_char);
while ((length < MAXCHARS) && (next_char != dollarchar))
{
OChars [length++] = next_char;
get_char (next_char);
}
}
//---------------------------------------------------------------------------------------------------------------
//----------------- ENCRYPTION ROUTINES -------------------------------------------------------------------------
void encrypt_chars (int length, char EKey)
{ char temp_char; // char temporary store
for (int i = 0; i < length; i++) // encrypt characters one at a time
{
temp_char = OChars [i]; //
__asm { //
push eax // save register values on stack to be safe
push ecx // Push last parameter first
lea eax,EKey
push temp_char
push eax
call encrypt3 // encrypt the character
mov temp_char, al
add esp, 8 // Clean parameters from stack
pop ecx // restore original register values from stack
pop eax //
}
EChars [i] = temp_char; // Store encrypted char in the encrypted chars array
}
return;
__asm {
encrypt3:
push ebp // Save the old base pointer value
mov ebp, esp // Set the new base pointer value
push edx // Save EDX to the first unused empty stack
push ecx //ecx register containing the temp_char is pushed to the stack
push eax // Save EAX to the first unused empty stack
mov edx, [ebp + 8] // Accessing the last value of ebp
movzx eax, byte ptr[eax] // Move 4 bytes to the EAX register
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
mov edx, eax // Move 4 bytes from EAX into edx
pop eax // Restore original EAX
mov byte ptr[eax], dl //moves the Ekey value into the EAX register as an 8-bit value
pop ecx //stores the current letter being encrypted within the ECX register (it was pushed to the stack earlier in the assembly code).
xor ecx, edx //clears the EDX register of all values
mov eax, ecx // Move 4 bytes from ECX into EAX
ror al, 1 // Rotate AL register one position to the left
ror al, 1 // Rotate AL register one position to the left
ror al, 1 // Rotate AL register one position to the left
pop edx // Restore the value of EDX
pop ebx // Restore original EBX
mov esp, ebp // Dellocate local variables
pop ebp // Restore the original value of EBP
ret // Return EAX value
}
//--- End of Assembly code
}
// end of encrypt_chars function
//---------------------------------------------------------------------------------------------------------------
//---------------------------------------------------------------------------------------------------------------
//----------------- DECRYPTION ROUTINES -------------------------------------------------------------------------
//
void decrypt_chars(int length, char EKey)
{
return;
}
// end of decrypt_chars function
//---------------------------------------------------------------------------------------------------------------
答案 0 :(得分：1)
当然，在放入汇编语言之前，请用高级语言编写代码。
以下是一些原因：
edx注册mov edx, [ebp + 8] // Accessing the last value of ebp
movzx eax, byte ptr[eax] // Move 4 bytes to the EAX register
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
mov edx, eax // Move 4 bytes from EAX into edx
在上面的代码中，您将[ebp + 8]移至edx。然后，您稍后将eax复制到edx四条说明中。为什么要在这里打扰第一条指令？
汇编语言编码的一个常见原因是效率。据说你可以比编译器更好地编码或者比编译器更好地使用特殊指令。你没有，正如这个例子所示：
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
rol al, 1 // Rotate AL register one position to the left
上述内容应编码为rol al, 3
另外，您是否有理由使用al注册代替eax？
清除edx个注册表是错误的
该操作与注释不匹配。
xor ecx, edx //clears the EDX register of all values
edx注册表是错误的
语句xor edx, edx实际上清除了edx寄存器。
我建议废弃汇编语言并用高级语言重写函数。让它先工作。检查编译器的汇编语言。如果您可以比编译器更有效地编写算法代码，那么就这样做。