c++ - 模板类的不同对象的通用接口(interface)

Question

C++ 新手，遇到过这样的代码(也不确定这是否是正确的做法)

template<ReqType, RespType>
class StreamManager {
...
...
}

using UpstreamManager = StreamManager<UpstreamReq, UpstreamResp>;
using DownstreamManager = StreamManager<DownstreamReq, DownstreamResp>;
using SidestreamManager = StreamManager<SidestreamReq, SidestreamResp>;

我想实现一个函数，它根据传递给函数的类型返回这些类型之一。有没有办法将其定义为某种常见类型。

**[ReturnType]** CreateStreamManager(StreamType type) {
    switch(type) {
        case upstream:
            return new UpstreamManager();
        case downstream:
            return new DownstreamManager();
        case sidestream:
            return new SidestreamManager();
    }
}

有没有办法为这个 CreateStreamManager 函数定义这个 ReturnType ？

Answer 1

您可以像这样“重载”函数:

UpstreamManager* CreateStreamManager(UpstreamReq req) {
    return new UpstreamManager();
}
DownstreamManager* CreateStreamManager(DownstreamReq req) {
    return new DownstreamManager();
}

或者您可以编写一个“工厂函数”，但前提是 StreamManager 继承了一个可以是通用返回类型的非模板基类:

class BaseStreamManager {
    virtual ~BaseStreamManager() = default;
};

template<ReqType, RespType>
class StreamManager : public BaseStreamManager {
}

BaseStreamManager* CreateStreamManager(StreamType type) {
    switch(type) {
        case upstream:
            return new UpstreamManager();
        case downstream:
            return new DownstreamManager();
        case sidestream:
            return new SidestreamManager();
    }
}