对于模板化类的内部模板化结构，我想要一个可变参数模板化构造函数。不幸的是，构造函数（参见下面的第一个构造函数）是不够的：如果我只使用那个构造函数，我会得到 C2260 编译器错误，指出构造函数不接受 3、4 或 5 个参数。另一方面，通过添加其他三个构造函数（请参阅下面的其余构造函数）来使所有内容都明确可按预期工作。
template< typename KeyT, typename ResourceT >
class ResourcePool {
...
template< typename DerivedResourceT >
struct ResourcePoolEntry final : public DerivedResourceT {
template< typename... ConstructorArgsT >
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ConstructorArgsT... args)
: DerivedResourceT(args...), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ID3D11Device2 &x)
: DerivedResourceT(x), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ID3D11Device2 &x, const wstring &y)
: DerivedResourceT(x,y), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
template < typename VertexT >
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ID3D11Device2 &x, const wstring &y, const MeshDescriptor< VertexT > &z)
: DerivedResourceT(x, y, z), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
...
}
}
构造函数是这样调用的：
template< typename KeyT, typename ResourceT >
template< typename... ConstructorArgsT >
std::shared_ptr< ResourceT > ResourcePool< KeyT, ResourceT >::GetResource(KeyT key, ConstructorArgsT... args) {
return GetDerivedResource< ResourceT, ConstructorArgsT... >(key, args...);
}
template< typename KeyT, typename ResourceT >
template< typename DerivedResourceT, typename... ConstructorArgsT >
std::shared_ptr< ResourceT > ResourcePool< KeyT, ResourceT >::GetDerivedResource(KeyT key, ConstructorArgsT... args) {
...
auto new_resource = std::shared_ptr< ResourcePoolEntry< DerivedResourceT > >(
new ResourcePoolEntry< DerivedResourceT >(*this, key, args...));
...
}
对于像boolas 可变参数这样的原语，一切正常。
Severity Code Description Project File Line Suppression State
Error C2660 'mage::ResourcePool<std::wstring,mage::VertexShader>::ResourcePoolEntry<DerivedResourceT>::ResourcePoolEntry': function does not take 3 arguments MAGE c:\users\matthias\documents\visual studio 2015\projects\mage\mage\mage\src\resource\resource_pool.tpp 37
其中第 37 行对应于构造函数调用（new ResourcePoolEntry< DerivedResourceT >(*this, key, args...));在上面的示例中）
我究竟做错了什么？（编译器 MSVC++ 14.0）
#include <memory>
#include <map>
template < typename T >
using SharedPtr = std::shared_ptr< T >;
template < typename T >
using WeakPtr = std::weak_ptr< T >;
template< typename KeyT, typename ResourceT >
using ResourceMap = std::map< KeyT, WeakPtr< ResourceT > >;
template< typename KeyT, typename ResourceT >
class ResourcePool {
public:
template< typename... ConstructorArgsT >
SharedPtr< ResourceT > GetResource(KeyT key, ConstructorArgsT... args);
template< typename DerivedResourceT, typename... ConstructorArgsT >
SharedPtr< ResourceT > GetDerivedResource(KeyT key, ConstructorArgsT... args);
private:
ResourceMap< KeyT, ResourceT > m_resource_map;
template< typename DerivedResourceT >
struct ResourcePoolEntry final : public DerivedResourceT {
public:
template< typename... ConstructorArgsT >
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ConstructorArgsT... args)
: DerivedResourceT(args...), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
private:
ResourcePool< KeyT, ResourceT > &m_resource_pool;
KeyT m_resource_key;
};
};
template< typename KeyT, typename ResourceT >
template< typename... ConstructorArgsT >
SharedPtr< ResourceT > ResourcePool< KeyT, ResourceT >::GetResource(KeyT key, ConstructorArgsT... args) {
return GetDerivedResource< ResourceT, ConstructorArgsT... >(key, args...);
}
template< typename KeyT, typename ResourceT >
template< typename DerivedResourceT, typename... ConstructorArgsT >
SharedPtr< ResourceT > ResourcePool< KeyT, ResourceT >::GetDerivedResource(KeyT key, ConstructorArgsT... args) {
auto it = m_resource_map.find(key);
if (it != m_resource_map.end()) {
auto resource = it->second.lock();
if (resource) {
return resource;
}
else {
m_resource_map.erase(it);
}
}
auto new_resource = SharedPtr< ResourcePoolEntry< DerivedResourceT > >(
new ResourcePoolEntry< DerivedResourceT >(*this, key, args...));
m_resource_map[key] = new_resource;
return new_resource;
}
#include <d3d11_2.h>
struct A {
};
struct B : public A {
B(ID3D11Device &device) : A() {}
};
const D3D_FEATURE_LEVEL g_feature_levels[] = {
D3D_FEATURE_LEVEL_11_1,
D3D_FEATURE_LEVEL_11_0
};
int main() {
ID3D11Device *device;
ID3D11DeviceContext *device_context;
D3D_FEATURE_LEVEL feature_level;
D3D11CreateDevice(nullptr, D3D_DRIVER_TYPE_HARDWARE, nullptr, 0,
g_feature_levels, _countof(g_feature_levels), D3D11_SDK_VERSION,
&device, &feature_level, &device_context
);
ResourcePool< char, A > *pool = new ResourcePool< char, A >();
//pool->template GetResource< int & >('a');
pool->template GetDerivedResource< B, ID3D11Device & >('b', *device);
}
错误：
Severity Code Description Line Suppression State
Error C2661 'ResourcePool<char,A>::ResourcePoolEntry<DerivedResourceT>::ResourcePoolEntry': no overloaded function takes 3 arguments 66
需要注意的一件事（这可能是也可能不是问题的最终原因）是您在模板参数转发方面做得并不正确。例如，在情况下你传递一个ID3D11Device2你DerivedResourceT可能构造函数（由非可变参数的构造函数的署名判定）预计参考-但是，因为模板的方式演绎的作品，它实际上会得到一个副本，而不是（如果确实此甚至是允许的 - 如果不是它就不会编译）。
要纠正此问题，您需要使用标准转发配方，它允许传递传递参数的正确 l 或 r 值，其中包括正确转发引用：
template< typename... ConstructorArgsT >
ResourcePoolEntry(ResourcePool< KeyT, ResourceT > &resource_pool,
KeyT resource_key, ConstructorArgsT&&... args)
: DerivedResourceT(std::forward<ConstructorArgsT>(args)...), m_resource_pool(resource_pool), m_resource_key(resource_key) {}
在上面，注意&&在参数类型args...和std::forward调用中。
本文收集自互联网，转载请注明来源。
如有侵权，请联系 [email protected] 删除。
我来说两句