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Home/ Questions/Q 8830687
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T08:09:07+00:00 2026-06-14T08:09:07+00:00

I’m currently working on a UDP server. I want to redirect all incoming packets

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I’m currently working on a UDP server.
I want to redirect all incoming packets to the connected clients by using the ip address and port.
My current way of doing it looks like this:

class Connection;
typedef std::map<unsigned short, Connection*> PortMap;
typedef std::map<unsigned int, PortMap> AddressMap;

So I’m basically using two maps. The second one contains a map of all the ports using an ipv4 address(unsigned int) as a key. The PortMap uses the port as the key and it contains a pointer to the Connection class(the clients).

I speed tested it by accessing 64 clients using randomly generated ips and ports and it took ~ (EDIT : 0.4 milliseconds) to access 64 different clients 64 times.
I don’t know really if it’s slow or not. Of course it depends on the system I’m running the test on.

Here’s how I’m accessing the client using the address:

Client * GetClient(Address address)
{
    AddressMap::iterator ipIt;
    PortMap::iterator portIt;
    unsigned int ip = address.GetAddress();
    unsigned short port = address.GetPort();

    /// Does the ip exist?
    if((ipIt = clientAddresses.find(ip)) == clientAddresses.end())
    {
        return NULL;
    }

    /// Does the port exist?
    if(clientAddresses[ip].find(port) == clientAddresses[ip].end())
    {   
        return NULL;
    }

    return clientAddresses[ip][port];
}

Does anyone know another faster way of doing it?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T08:09:09+00:00Added an answer on June 14, 2026 at 8:09 am

    Maybe it’ll be better to combine IP and Port..

    Port < 65535. You can get key: IP * 65535 + Port, it’ll be unique for all ports and IPs.

    About speed: for example we have N Ip’s, each IP has M ports.

    Searching into map has efficient N log(N), so your search take N*M*log(N)*log(M).
    If you combine IP and port into one key, efficient will be N*M*log(N*M).
    log(N*M) = log(N)+log(M) < log(N)*log(M), for big N, M..

    So, I think it’ll be better.

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