複雜的C++模板代碼

Question

我一直在爲opencascade C++封裝到C＃。 我的C++有點生疏，因爲我最近幾年主要使用C＃。

現在我遇到了以下問題，一直無法弄清楚如何糾正它：

#include <type_traits> 

//! Trait yielding true if class T1 is base of T2 but not the same 
template <class T1, class T2, class Dummy = void> 
struct is_base_but_not_same : std::is_base_of <T1, T2> {}; 

//! Explicit specialization of is_base_of trait to workaround the 
//! requirement of type to be complete when T1 and T2 are the same. 
template <class T1, class T2> 
struct is_base_but_not_same <T1, T2, typename std::enable_if <std::is_same <T1, T2>::value>::type> : std::false_type {}; 

template <class T> 
class handle 
{ 
public: 
    //! Down casting operator from handle to base type 
    template <class T2> 
    static typename std::enable_if<is_base_but_not_same<T2, T>::value, handle>::type 
     DownCast(const handle<T2>& theObject) 
    { 
     return handle(dynamic_cast<T*>(const_cast<T2*>(theObject.get()))); 
    } 

    //! Down casting operator from pointer to base type 
    template <class T2> 
    static typename std::enable_if<is_base_but_not_same<T2, T>::value, handle>::type 
     DownCast(const T2* thePtr) 
    { 
     return handle(dynamic_cast<T*>(const_cast<T2*>(thePtr))); 
    } 
}; 

class Foo 
{ 

}; 

typedef handle<Foo> Handle_Foo; 

Handle_Foo DownCastFoo(Handle_Foo const &T) { 
    return Handle_Foo::DownCast(T); 
} 

的compilor錯誤有以下幾種：

Error C2672 'handle<Foo>::DownCast': no matching overloaded function found 
Severity Code Description Project File Line Suppression State 
Error C2784 'std::enable_if<is_base_but_not_same<T2,T,void>::value,handle<T>>::type handle<T>::DownCast(const T2 *)': could not deduce template argument for 'const T2 *' from 'const Handle_Foo' 
Severity Code Description Project File Line Suppression State 
Error C2893 Failed to specialize function template 'std::enable_if<is_base_but_not_same<T2,T,void>::value,handle<T>>::type handle<T>::DownCast(const handle<T2> &)' 

任何人都可以點我在正確的方向？

Answer 1

在提供的示例中有多個錯誤。

首先，例如意味着對於handle類T*構造和get()方法。這些添加到handle類：

explicit handle(T*); 
const T* get() const; 

其次，你正試圖從handle<Foo>下來投給handle<Foo>這是一場毫無意義的操作。 Foo甚至不是多態的。你的下調方法似乎專門設計爲不提供這種超載，因此你得到的錯誤。將Foo定義爲多態，並添加派生類Bar。

struct Foo { virtual ~Foo() = default; }; 
struct Bar : public Foo {}; 

最後，改變DownCastFoo嘗試向下轉換Bar而不是Foo，並且該錯誤被解決。

typedef handle<Foo> Handle_Foo; 
typedef handle<Bar> Handle_Bar; 

Handle_Bar DownCastFoo(Handle_Foo const &T) { 
    return Handle_Bar::DownCast(T); 
} 

最後一點，那些const_cast氣色好可疑。不幸的是，您的示例沒有足夠的信息來提供更好的建議。

---

完全正確的示例

#include <type_traits> 

//! Trait yielding true if class T1 is base of T2 but not the same 
template <class T1, class T2, class Dummy = void> 
struct is_base_but_not_same : std::is_base_of <T1, T2> {}; 

//! Explicit specialization of is_base_of trait to workaround the 
//! requirement of type to be complete when T1 and T2 are the same. 
template <class T1, class T2> 
struct is_base_but_not_same <T1, T2, typename std::enable_if <std::is_same <T1, T2>::value>::type> : std::false_type {}; 

template <class T> 
class handle 
{ 
public: 
    explicit handle(T*); 
    const T* get() const; 
public: 
    //! Down casting operator from handle to base type 
    template <class T2> 
    static typename std::enable_if<is_base_but_not_same<T2, T>::value, handle>::type 
     DownCast(const handle<T2>& theObject) 
    { 
     return handle(dynamic_cast<T*>(const_cast<T2*>(theObject.get()))); 
    } 

    //! Down casting operator from pointer to base type 
    template <class T2> 
    static typename std::enable_if<is_base_but_not_same<T2, T>::value, handle>::type 
     DownCast(const T2* thePtr) 
    { 
     return handle(dynamic_cast<T*>(const_cast<T2*>(thePtr))); 
    } 
}; 

struct Foo { virtual ~Foo() = default; }; 
struct Bar : public Foo {}; 

typedef handle<Foo> Handle_Foo; 
typedef handle<Bar> Handle_Bar; 

Handle_Bar DownCastFoo(Handle_Foo const &T) { 
    return Handle_Bar::DownCast(T); 
} 

---

編輯：它看起來像enable_if僅用於防止非法超載。如果您使用static_assert，您將得到更好的編譯錯誤。

示例使用static_assert：

template <class T> 
class handle 
{ 
public: 
    explicit handle(T*); 
    const T* get() const; 
public: 
    //! Down casting operator from handle to base type 
    template <class T2> 
    static handle DownCast(const handle<T2>& theObject) 
    { 
     static_assert(std::is_same<T, T2>::value == false, 
      "Can't downcast from between identical types"); 
     static_assert(std::is_base_of<T2, T>::value, 
      "Can only down cast from a derived type to a base type"); 
     return handle(dynamic_cast<T*>(const_cast<T2*>(theObject.get()))); 
    } 

    //! Down casting operator from pointer to base type 
    template <class T2> 
    static handle DownCast(const T2* thePtr) 
    { 
     static_assert(std::is_same<T, T2>::value == false, 
      "Can't downcast from between identical types"); 
     static_assert(std::is_base_of<T2, T>::value, 
      "Can only down cast from a derived type to a base type"); 
     return handle(dynamic_cast<T*>(const_cast<T2*>(thePtr))); 
    } 
}; 

現在，您原來使用的情況下會產生錯誤信息"Can't downcast from between identical types"。

struct Foo { virtual ~Foo() = default; }; 
struct Bar : public Foo {}; 

typedef handle<Foo> Handle_Foo; 

Handle_Foo DownCastFoo(Handle_Foo const &T) { 
    return Handle_Foo::DownCast(T); 
}