2015-09-14 63 views
1

我试图构建zguide C++示例,但编译失败。以下是错误输出中:zGuide for zeroMQ编译失败

asyncsrv.cpp: In member function ‘void client_task::start()’: 
asyncsrv.cpp:37:70: error: cannot convert ‘zmq::socket_t’ to ‘void*’ in initialization zmq::pollitem_t items[] = {{client_socket_, 0, ZMQ_POLLIN, 0}}; 

asyncsrv.cpp: In member function ‘void server_task::run()’: asyncsrv.cpp:140:52: error: cannot convert ‘zmq::socket_t’ to ‘void*’ for argument ‘1’ to ‘void zmq::proxy(void*, void*, void*)’ zmq::proxy(frontend_, backend_, nullptr); 
+0

听起来可能是您安装的ZMQ和您试图运行的示例代码之间的版本不兼容...您介意给出您的ZMQ版本和示例代码的链接? – Jason

+0

zmqversion是4.1.3,zGuide具有最新的标签v2.2-1024-g40230a3并提交sha1 40230a37eb45d8564733bedc08c333190387df3d。 C样本构建得很好。 –

回答

8

如果您使用的是C++编译器11,看到你的zmq.hpp是最新版本,并通过

std::vector<zmq::pollitem_t> items = {{static_cast<void *>(client_socket_), 0, ZMQ_POLLIN, 0}}; 
更换线

zmq::pollitem_t items[] = {{client_socket_, 0, ZMQ_POLLIN, 0}}; 

and line

zmq::proxy(frontend_, backend_, nullptr); 

by

zmq::proxy(static_cast<void *>(frontend_), static_cast<void *>(backend_), nullptr); 
+0

我在gcc 5上遇到了这个问题,而它在铿锵声中编译得很好 – fferri