运行 shellcode 存储在动态允许的内存中

Question

我有以下运行可以正常工作的 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 个不同数组中的相同 shellcode 时，我遇到了访问冲突：

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)();

我正在尝试使第二个示例正常运行，但我遇到了访问冲突错误。我究竟做错了什么？谢谢。

更新

根据 alain 和三十二上校的建议修改后的代码，我现在得到以下错误："test.exe has triggered a breakpointt"

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

一个字符串文字以空字符结尾，并且以 sizeof 计算终止的空字节。因此，当使用 2 数组版本时，最终数组的中间有一个空字节。

如果你改变

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

到

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

我觉得应该可以。

正如三十二上校所指出的，sizeof concatenation也有错误。

运行 shellcode 存储在动态允许的内存中

run shellcode stored in dynamically allowed memory

c

memory

windows

shellcode