ZeroMQ多部分消息第二部分0尺寸

Question

我想在ZMQ 4.2.3和cppzmq中使用子/ pub模式接收多部分消息我能夠成功發送和接收單個部分消息，但是當我嘗試在第二幀中讀取它的大小始終爲0.我完全困惑的是使用NetMQ的C＃版本讀取第二幀沒有問題，導致我相信它正在正確發送。我知道我錯過了一些東西，但今天是我試圖找出沒有成功的第二天。

這裏是我的酒吧代碼

#include <iostream> 
#include "zmq_addon.hpp" 
void main() 
{ 
    zmq::context_t ctx = zmq::context_t(); 
    zmq::socket_t pub = zmq::socket_t(ctx, zmq::socket_type::pub); 
    try 
    { 
     //binding using localhost gives an error about invalid device? 
     pub.bind("tcp://*:8845"); 
    } 
    catch (...) 
    { 
     std::cout << zmq_strerror(zmq_errno()); 
     std::cin.get(); 
     return; 
    } 
    byte topic = 8; 
    std::string id = "ragefire.bob"; 
    while (true) 
    { 
     std::cout << "Spam\n"; 
     pub.send(id.c_str(), id.length(), ZMQ_SNDMORE); 
     pub.send("foo", 3); 
    } 
} 

而我的C++子代碼

#include <iostream> 
#include "zmq_addon.hpp" 
int main() 
{ 
    zmq::context_t ctx = zmq::context_t(); 
    zmq::socket_t sub = zmq::socket_t(ctx, zmq::socket_type::sub); 
    sub.connect("tcp://localhost:8845"); 
    std::string id = "ragefire.bob"; 
    sub.setsockopt(ZMQ_SUBSCRIBE, id.c_str(), id.length()); 
    while (true) 
    { 
     zmq::message_t msg; 
     if(sub.recv(&msg,ZMQ_NOBLOCK)) 
     { 
      auto rpl = std::string(static_cast<char*>(msg.data()), msg.size()); 
      std::cout << "Recv returned true! " << rpl << "\n"; 
      int more; 
      auto more_size = sizeof(more); 
      sub.getsockopt(ZMQ_RCVMORE, &more, &more_size); 
      while (more) 
      { 
       zmq::message_t moreMsg; 
       sub.recv(&msg); 
       std::string moreRpl = std::string(static_cast<char*>(moreMsg.data()), moreMsg.size()); 
       std::cout << "There's more! " << moreRpl << "Size is " << moreMsg.size() << "\n"; 
       sub.getsockopt(ZMQ_RCVMORE, &more, &more_size); 
      } 
     } 
     else 
      std::cout << "Recv returned false! " << zmq_strerror(zmq_errno()) << "\n"; 
    } 
} 

隨着

Recv returned true! ragefire.bob 
There's more! Size is 0 

輸出爲了完整起見，我NetMQ子，可以讀取兩個框架

static void Main() 
{ 
    SubscriberSocket sub = new SubscriberSocket("tcp://localhost:8845"); 
    sub.SubscribeToAnyTopic(); 
    while (true) 
    { 
     NetMQMessage msg = sub.ReceiveMultipartMessage(); 
     Console.WriteLine($"Received! {Encoding.ASCII.GetString(msg.First.Buffer)}"); 
     Console.WriteLine($"Received! {Encoding.ASCII.GetString(msg[1].Buffer)}"); 
    } 
} 

Answer 1

您對sub.recv()初始呼叫中&msg接收到該消息，這是好的。

您隨後撥打sub.recv()也會收到&msg，但您隨後會報告有關moreMsg。如果你真的想保持第二message_t左右，做

sub.recv(&moreMsg);