我真的很想弄清楚我已经接受过的这项任务，现在我已经不在了。
我得到了以下循环的示例：
#include <iostream> // for cout
using namespace std;
//declare an array of 'marks'.
const int array_size = 5;
int marks [] = {45,56,99,19,21};
int main()
{int index, average, total;
total = 0;
for ( index = 0; index < array_size; index++)
{
total = total + marks [index];
}
average = total/array_size;
cout << "\nAverage value = " << average << "\n\n";
return(0);
}
：
#include <iostream> // for cout
using namespace std;
const int array_size = 5;
int marks [] = {45,56,99,19,21}; //declare an array of 'marks'.
int main()
{int index, average, total;
__asm {
mov total,0 ;total = 0;
; ************************** This part is the FOR loop *************************
mov index,0 ;for ( index = 0; index < array_size; index++)
jmp checkend
forloop1:
mov eax,index ;add 1 to index
add eax,1
mov index,eax
checkend:
cmp index,5 ;check if 5 ('array_size') loops have been done
jge endfor1 ;jump if greater than or equal to (remember?)
mov ecx,index
mov edx,total ; total =
add edx,[ecx*4+marks] ; total + marks [index];
mov total,edx
jmp forloop1
; ******************************************************************************
endfor1:
mov eax,total ;average = total/array_size;
cdq ;convert EAX to quadword (uses EDX register)
mov ecx,5 ;get array_size as divisor
idiv ecx ;divides EDX:EAX pair by ECX, EAX=answer
mov average,eax ;save answer in variable 'average'
} //end of assembly section
cout << "\nAverage value = " << average << "\n\n";
return(0);
}
然而，我的问题是这样的：我需要将以下循环转换为汇编，根据我给出的示例，这样做的最佳方法是什么？我真的很挣扎着将temp_char变量赋给数组变量。
通过以下循环，我的意思是 encrypt_chars函数循环
#define MAXCHARS 6
#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);
}
}
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]; // temp_char now contains the address values of the individual character
__asm
{
push eax // Save values contained within register to stack
push ecx
movzx ecx, temp_char
push ecx // Push argument #2
lea eax, EKey
push eax // Push argument #1
call encrypt4
add esp, 8 // Clean parameters of stack
mov temp_char, al // Move the temp character into a register
pop ecx
pop eax
}
EChars[i] = temp_char; // Store encrypted char in the encrypted chars array
}
return;
// Inputs: register EAX = 32-bit address of Ekey,
// ECX = the character to be encrypted (in the low 8-bit field, CL).
// Output: register EAX = the encrypted value of the source character (in the low 8-bit field, AL).
__asm
{
encrypt4:
push ebp // Set stack
mov ebp, esp // Set up the base pointer
mov eax, [ebp + 8] // Move value of parameter 1 into EAX
mov ecx, [ebp + 12] // Move value of parameter 2 into ECX
push edi // Used for string and memory array copying
push ecx // Loop counter for pushing character onto stack
not byte ptr[eax] // Negation
add byte ptr[eax], 0x04 // Adds hex 4 to EKey
movzx edi, byte ptr[eax] // Moves value of EKey into EDI using zeroes
pop eax // Pop the character value from stack
xor eax, edi // XOR character to give encrypted value of source
pop edi // Pop original address of EDI from the stack
rol al, 1 // Rotates the encrypted value of source by 1 bit (left)
rol al, 1 // Rotates the encrypted value of source by 1 bit (left) again
add al, 0x04 // Adds hex 4 to encrypted value of source
mov esp, ebp // Deallocate values
pop ebp // Restore the base pointer
ret
}
//--- End of Assembly code
}
int main(void)
{
int char_count; // The number of actual characters entered (upto MAXCHARS limit).
char EKey; // Encryption key.
cout << "\nPlease enter your Encryption Key (EKey) letter: "; get_char(EKey);
cout << "\nNow enter upto " << MAXCHARS << " alphanumeric characters:\n";
get_original_chars(char_count);
cout << "\n\nOriginal source string = " << OChars << "\tHex = ";
for (int i = 0; i<char_count; i++) cout << hex << setw(2) << setfill('0') << ((int(OChars[i])) & 0xFF) << " ";
encrypt_chars(char_count, EKey);
cout << "\n\nEncrypted string = " << EChars << "\tHex = ";
for (int i = 0; i<char_count; i++) cout << ((int(EChars[i])) & 0xFF) << " ";
decrypt_chars(char_count, EKey);
cout << "\n\nDecrypted string = " << DChars << "\tHex = ";
for (int i = 0; i<char_count; i++) cout << ((int(DChars[i])) & 0xFF) << " ";
cout << "\n\nPress a key to end...";
while (!_kbhit()); //hold the screen until a key is pressed
return (0);
}
提供最佳解释（逐步）答案/解决方案的cookie !!
编辑：从循环等生成的反汇编代码：
答案 0 :(得分：0)
注意第32行，其中表示mov cl，byte ptr [eax + 0CC0320h]除非你 可以告诉我如何将其转换为正常的asm代码＆＃39;
mov cl,byte ptr [eax+0CC0320h]：eax此时包含i的值，0CC0320h是OChars的地址，即它的开头。所以，＆＃34;正常＆＃34; asm代码如下：
__asm {
mov eax, i
mov cl, byte ptr OChars[eax]
mov temp_char, cl
}