how to subscribe to keyspace changes and determine name of keys that changed?

[AlDanial]

I'm trying to write a program that subscribes to a keyspace change pattern. If it detects that a key has changed, I'd like to know which specific key it was. I began with redisclient-0.6.1/examples/async_pubsub2.cpp then simplified it to

#include <string>
#include <iostream>
#include <functional>
#include <boost/asio/ip/address.hpp>
#include <redisclient/redisasyncclient.h>

static const std::string channelName = "__keyspace@0__:x*";

class Client {
public:
    Client(boost::asio::io_service &ioService)
        : ioService(ioService) {}

    void onMessage(const std::vector<char> &buf) {
        std::string msg(buf.begin(), buf.end());
        std::cerr << "Message: " << msg << std::endl;
        if( msg == "stop" )
            ioService.stop();
    }

private:
    boost::asio::io_service &ioService;
};

int main(int, char **) {
    boost::asio::ip::address address = boost::asio::ip::address::from_string("127.0.0.1");
    const unsigned short port = 6379;
    boost::asio::ip::tcp::endpoint endpoint(address, port);

    boost::asio::io_service ioService;
    redisclient::RedisAsyncClient subscriber(ioService);
    Client client(ioService);

    subscriber.connect(endpoint, [&](boost::system::error_code ec) {
        if( ec ) {
            std::cerr << "Can't connect to redis: " << ec.message() << std::endl;
        } else {
            subscriber.psubscribe(channelName,
                    std::bind(&Client::onMessage, &client, std::placeholders::_1)
                    );
        }
    });
    ioService.run();
    return 0;
}

which subscribes to changes to any key beginning with x. It half-works: when I change the value of key xy (manually, using redis-cli from another terminal), the onMessage() method in the program above is called, but msg just contains "set". But which key was set--xy or xz or .... ?

How should the code be changed so that it also captures the name(s) of the changed key(s)?

[nekipelov]

Currently there is no way. I'll do it later.