如何准备 msgpack 以通过 ZeroMQ 基础结构发送结构?
How to prepare msgpack to send a struct over a ZeroMQ infrastructure?
我正在尝试通过 ZeroMQ 在节点之间进行通信。我需要发送 struct
个:
struct Content{
std::vector<cv::Mat> image;
std::string msg;
};
我尝试利用 msgpack
:
Content content;
content.image = msg2;
content.mesaj = "naber kardes";
msgpack::type::tuple<Content> src(content);
// serialize the object into the buffer.
// any classes that implements
// write(const char*,size_t) can be a buffer.
std::stringstream buffer;
msgpack::pack(buffer, src);
cout << sizeof(buffer) << endl;
但它给出:
/usr/local/include/msgpack/v1/object.hpp:631:11: error: no member named 'msgpack_pack' in 'Content'
对 c++ 还是新手。
如何在 msgpack
的帮助下通过 ZeroMQ 发送我的内容结构?
您需要为您的 Content
结构提供 pack()
函数:
namespace msgpack
{
MSGPACK_API_VERSION_NAMESPACE(MSGPACK_DEFAULT_API_NS)
{
namespace adaptor
{
template <>
struct pack<Content> {
template <typename Stream>
msgpack::packer<Stream>& operator()(
msgpack::packer<Stream>& out, Content const& obj) const
{
out.pack(obj.image);
out.pack(obj.msg);
return out;
}
};
}
}
}
如果您的软件不评估 MessagePack 数据,您需要根据 zmq/MessagePack 的 format/layout 期望创建 pack() 函数收件人。
convert() 函数示例(与上述 pack() 函数相同的命名空间):
template <>
struct convert<Content> {
msgpack::object const& operator()(
msgpack::object const& o, Content& v) const
{
// unpack data in the same format as it was packed. see above!
return o;
}
};
我正在尝试通过 ZeroMQ 在节点之间进行通信。我需要发送 struct
个:
struct Content{
std::vector<cv::Mat> image;
std::string msg;
};
我尝试利用 msgpack
:
Content content;
content.image = msg2;
content.mesaj = "naber kardes";
msgpack::type::tuple<Content> src(content);
// serialize the object into the buffer.
// any classes that implements
// write(const char*,size_t) can be a buffer.
std::stringstream buffer;
msgpack::pack(buffer, src);
cout << sizeof(buffer) << endl;
但它给出:
/usr/local/include/msgpack/v1/object.hpp:631:11: error: no member named 'msgpack_pack' in 'Content'
对 c++ 还是新手。
如何在 msgpack
的帮助下通过 ZeroMQ 发送我的内容结构?
您需要为您的 Content
结构提供 pack()
函数:
namespace msgpack
{
MSGPACK_API_VERSION_NAMESPACE(MSGPACK_DEFAULT_API_NS)
{
namespace adaptor
{
template <>
struct pack<Content> {
template <typename Stream>
msgpack::packer<Stream>& operator()(
msgpack::packer<Stream>& out, Content const& obj) const
{
out.pack(obj.image);
out.pack(obj.msg);
return out;
}
};
}
}
}
如果您的软件不评估 MessagePack 数据,您需要根据 zmq/MessagePack 的 format/layout 期望创建 pack() 函数收件人。
convert() 函数示例(与上述 pack() 函数相同的命名空间):
template <>
struct convert<Content> {
msgpack::object const& operator()(
msgpack::object const& o, Content& v) const
{
// unpack data in the same format as it was packed. see above!
return o;
}
};