Question

例如，使用以下代码：

    function calculatingDistance(fromLat,fromLong,toLat,toLong){
    var distanceDurationObject={};
var options = {
  host: 'maps.googleapis.com',
  path: '/maps/api/directions/json?origin='+fromLat+','+fromLong+'&destination='+toLat+','+toLong+'&key=AIzaSyBsoc1PZOItqHNYc5z2hW_ejPFi2piRR8Y',
  method: 'GET',
    headers: {
        'Content-Type': 'application/json'
    } 
};

var req = https.get(options, function(res) {
 //buffering alll data
  var bodyChunks = [];
  res.on('data', function(chunk) {
    // You can process streamed parts here...
    bodyChunks.push(chunk);
  }).on('end', function() {
    var body = Buffer.concat(bodyChunks);
     json = JSON.parse(body);
    //parsing json returned
    var routes =json["routes"];
    var legsObject=routes[0].legs;
    var legsFirstArray=legsObject[0];
    //Our Required Distance between two points
    var distanceValue=legsFirstArray.distance.value;
    var durationValue=legsFirstArray.duration.value;
    console.log(distanceValue+"  "+durationValue);
    distanceDurationObject.distance=distanceValue;
    distanceDurationObject.duration=durationValue;

  });
});

req.on('error', function(e) {
  console.log('ERROR: ' + e.message);
  //try to add this error to a file
});
}

请注意＆＃39; Quux＆＃39; class有两个模板参数 - 一个也是＆＃39; Bar＆＃39;的模板参数。 class，以及对// Say I have this class defined in some other file class Foo; // This class will act as a wrapper for an integer map // to function pointers, which will create type TFoo objects // depending on the given input (in this case a "const char*" template<class TFoo> struct Bar { typedef TFoo foo_t; typedef TFoo (*get_foo_f_t)(const char*); typedef std::unordered_map<int, get_foo_f_t> foo_handler_map_t; Bar(const foo_handler_map_t& handlers) : handlers_(handlers) { } ~Bar() { } const foo_handler_map_t& handlers_; }; // Now, this class will receive an _object_ of type // "const Bar<T>&", which will have an already initialized // map of integers to function pointers, different // functions will be called with different input values // via the public method, "action()". template<class TFoo, const Bar<TFoo>& CBar> class Quux { public: Quux() : bar_(CBar) { } ~Quux() { } TFoo action(int a, const char* x) { auto it = this->bar_.handlers_.find(a); if (it == this->bar_.handlers_.end()) { // no handler defined for int `a' return TFoo(); } // i.e. CBar.handlers_[a](x) return it->second(x); } private: const Bar<TFoo>& bar_; }; // Here is how the map of integers to function pointers // will be initialized... static std::unordered_map<int, Foo (*)(const char*)> handlers { { 0, _hdl_0 }, // _hdl_* functions defined in different file { 1, _hdl_1 }, { 2, _hdl_2 } }; // And then passed to a "const Bar<T>" type object here const Bar<Foo> bar (handlers); int main() { // --> HERE IS WHAT I WANT TO CHANGE <-- Quux<decltype(bar)::foo_t, bar> quux; // ------------------------------------- // Example (trivial) use of the 'quux' object std::cout << quux.action(0, "abc").baz() << std::endl; std::cout << quux.action(1, "def").baz() << std::endl; std::cout << quux.action(2, "ghi").baz() << std::endl; return 0; }类型的模板对象的引用，其中T是与＆＃39; Foo＆＃39;相关的任何类。我希望能够做到以下几点：

const Bar<T>

注意：＆＃39; bar＆＃39;是Quux<bar> quux;类型的对象，但它也应该是任何Bar<Foo>类型。

这可能吗？我当时认为下面的内容可能会被用作快速解决方法，但我无法弄清楚应该用什么代替Bar<T>：

/* ??? */

修改的

我将对象的引用传递给了Quux＆＃39;作为模板参数，因为复制效率低（我认为），而不是复制整个template<const Bar</* ??? */>& CBar> using Nuff = Quux<decltype(CBar)::foo_t, CBar> Nuff<bar> nuff;对象。我只希望能够拥有一堆foo_handler_map_t类型的对象，这些对象在某个命名空间中全局定义，并且能够初始化＆＃39; Quux＆＃39;像这样的对象：

const Bar<T>

...而且我不想将它作为构造函数参数传递。

Answer 1

评论非常有用。但是，如果您希望减少模板参数的数量，可以将CBar作为参数传递给构造函数：

template<class TFoo>
class Quux
{
public:
    Quux(const Bar<TFoo>& CBar)
        : bar_(CBar)
    {}

    ~Quux()
    {}

    TFoo action(int a, const char* x)
    {
        auto it = this->bar_.handlers_.find(a);
        if (it == this->bar_.handlers_.end())
        {
        return TFoo();
        }
        return it->second(x);
    }

private:
    const Bar<TFoo>& bar_;
};

定义一个函数来创建Quux的实例：

template <typename TFoo>
auto make_Quux(const Bar<TFoo>& bar)
{
    return Quux<TFoo>(bar);
}

然后在main()中，您可以使用make_Quux()：

int main()
{
    auto quux = make_Quux(bar);
    //...
}

如何在没有底层模板参数的情况下将模板模板参数传递给模板类？

1 个答案: