最小示例：

Question

对于模板化类的内部模板化结构，我希望有一个可变的模板化构造函数。不幸的是，构造函数（参见下面的第一个构造函数）是不够的：如果我只使用该构造函数，我会获得C2260编译器错误，指出构造函数不带有3,4或5个参数。另一方面，通过添加另外三个构造函数（参见下面的剩余构造函数）使一切明确，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) {}

           ...
    }
}

构造函数的调用方式如下：

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bool这样的原语作为可变参数，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

第37行对应于构造函数调用（上例中的new ResourcePoolEntry< DerivedResourceT >(*this, key, args...));）

我做错了什么？（编译器MSVC ++ 14.0）

最小示例：

#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

Answer 1

需要注意的一件事（可能是或可能不是您问题的最终原因）是您在模板参数转发方面没有做正确的事情。例如，在您传递ID3D11Device2 DerivedResourceT构造函数的情况下（通过非变量构造函数的签名判断）需要引用 - 但是由于模板扣除的方式工作它实际上会得到一个副本（如果确实这甚至是允许的 - 如果不是它不会编译）。

要解决此问题，您需要使用标准转发配方，该配方允许传递传递参数的正确l值或r值，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) {}

在上文中，请注意参数类型&&和args...来电中的std::forward。

Variadic模板化构造函数不接受x参数

最小示例：

1 个答案: