将双精度转换为char数组而无需在C ++中进行填充

Question

I am trying to send a double through a UDP winsock using the Sendto function but first I need to convert it to a char[]. After trying multiple approaches, the best thing I came up with is the following:

    // send 
    memset(buf, '\0', BUFLEN);
    double a = position.y();
    char arr[sizeof(a)];
    cout << "double" << a << endl;
    ToString(arr, sizeof(a), a);

    sendto(s, arr, strlen(arr), 0, (struct sockaddr *) &si_other, slen);
    cout << "Send" << arr << endl;

This unfortunately gives out a weird trailing padding that ruins the send operation. Any ideas on how to solve this?

My ToString function is the following:

void ToString(char * outStr, int length, double val)
{
     _snprintf(outStr, length, "%f", val);
}

The output looks like this:

double 0.003
Send 0.003(a bunch of gibberish)

Answer 1

You are assuming that the number of bytes that the binary double occupies is the same length as the converted string occupies but that is unlikely.

If you have C++11 you can do something like this:

void send_double(int s, sockaddr* si_other, int slen, double d)
{
    auto buf = std::to_string(d);
    sendto(s, buf.data(), buf.size(), 0, si_other, slen);
}

Otherwise you can do something like this:

void send_double(int s, sockaddr* si_other, int slen, double d)
{
    std::ostringstream oss;
    oss << d;
    std::string buf = oss.str();
    sendto(s, buf.data(), buf.size(), 0, si_other, slen);
}

See std::string , std::to_string and std::ostringstream