C ++ WSAsend数据包的奇怪十六进制转储

Question

http://prntscr.com/2ctnoz

我正在挂钩WSAsend函数并转储数据包。 ASCII转储有效，但HEX转储有时会显示你可以在屏幕上看到的东西（FFFFFFDD），任何想法为什么？

码：

int WINAPI myWSASend(SOCKET s, LPWSABUF lpBuffers, DWORD dwBufferCount, LPDWORD lpNumberOfBytesSent, DWORD dwFlags, LPWSAOVERLAPPED lpOverlapped, LPWSAOVERLAPPED_COMPLETION_ROUTINE lpCompletionRoutine)
{   
    //Packet Log
    if (bLogPacketS == TRUE)
    {
        for (unsigned int i = 0; i < lpBuffers->len; i = i + 8)
        {
            printf("%02X %02X %02X %02X %02X %02X %02X %02X\t\t%c %c %c %c %c %c %c %c\n",
                  (unsigned int)lpBuffers->buf[i], (unsigned int)lpBuffers->buf[i+1], (unsigned int)lpBuffers->buf[i+2],
                  (unsigned int)lpBuffers->buf[i+3], (unsigned int)lpBuffers->buf[i+4], (unsigned int)lpBuffers->buf[i+5],
                  (unsigned int)lpBuffers->buf[i+6], (unsigned int)lpBuffers->buf[i+7],
                  (drawable((unsigned int)lpBuffers->buf[i])) ? (unsigned int)lpBuffers->buf[i] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+1])) ? (unsigned int)lpBuffers->buf[i+1] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+2])) ? (unsigned int)lpBuffers->buf[i+2] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+3])) ? (unsigned int)lpBuffers->buf[i+3] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+4])) ? (unsigned int)lpBuffers->buf[i+4] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+5])) ? (unsigned int)lpBuffers->buf[i+5] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+6])) ? (unsigned int)lpBuffers->buf[i+6] : '.',
                  (drawable((unsigned int)lpBuffers->buf[i+7])) ? (unsigned int)lpBuffers->buf[i+7] : '.');
        }
        printf("\n\n");
    }
    return (oWSASend)(s, lpBuffers, dwBufferCount, lpNumberOfBytesSent, dwFlags, lpOverlapped, lpCompletionRoutine);
}

bool drawable(unsigned int value)
{
    if (value > 32 && value < 127)
        return true;
    else
        return false;
}

Answer 1

你投的错了。

您只想更改签名，但您也将每个字节扩展为一个四字节的字。 在存在有符号到无符号的转换（以及由此产生的负值环绕）的情况下，这会导致极高的值。

---

lpBuffers->buf[i]每个“元素”都是一个char ，但是你要转换为unsigned int 。 如果您的char在您的系统上签名，则0xDD不在该类型范围的顶部，因此它将回绕到-35 。 然后将其转换为unsigned int导致0xFFFFFFDD 。

printf说明符%02X不会截断它。

假设您希望将所有字节解释为unsigned ，以获得完整的0x00 → 0xFF范围。 就个人而言，我将转换为unsigned char （而不是unsigned int ），其值为0xDD为221 。

在下面的代码中，我还对您的循环条件进行了安全调整。

for (unsigned int i = 0; i < lpBuffers->len-8; i = i + 8)
//                                         ^^
{
    printf(
        "%02X %02X %02X %02X %02X %02X %02X %02X"
          "\t\t%c %c %c %c %c %c %c %c\n",

        static_cast<unsigned char>(lpBuffers->buf[i]),
        static_cast<unsigned char>(lpBuffers->buf[i+1]),
        static_cast<unsigned char>(lpBuffers->buf[i+2]),
        static_cast<unsigned char>(lpBuffers->buf[i+3]),
        static_cast<unsigned char>(lpBuffers->buf[i+4]),
        static_cast<unsigned char>(lpBuffers->buf[i+5]),
        static_cast<unsigned char>(lpBuffers->buf[i+6]),
        static_cast<unsigned char>(lpBuffers->buf[i+7]),
        (drawable(lpBuffers->buf[i]))   ? static_cast<unsigned char>(lpBuffers->buf[i])   : '.',
        (drawable(lpBuffers->buf[i+1])) ? static_cast<unsigned char>(lpBuffers->buf[i+1]) : '.',
        (drawable(lpBuffers->buf[i+2])) ? static_cast<unsigned char>(lpBuffers->buf[i+2]) : '.',
        (drawable(lpBuffers->buf[i+3])) ? static_cast<unsigned char>(lpBuffers->buf[i+3]) : '.',
        (drawable(lpBuffers->buf[i+4])) ? static_cast<unsigned char>(lpBuffers->buf[i+4]) : '.',
        (drawable(lpBuffers->buf[i+5])) ? static_cast<unsigned char>(lpBuffers->buf[i+5]) : '.',
        (drawable(lpBuffers->buf[i+6])) ? static_cast<unsigned char>(lpBuffers->buf[i+6]) : '.',
        (drawable(lpBuffers->buf[i+7])) ? static_cast<unsigned char>(lpBuffers->buf[i+7]) : '.'
    );
}