使用SFINAE来检测C ++中的类型的POD [英] Using SFINAE to detect POD-ness of a type in C++

问题描述

这里的原始标题是
VS2005 C ++中的SFINAE错误的解决方法

这是暂时使用SFINAE等效于在TR1中存在的is_pod模板类（在VS2005中还没有TR1）。当模板参数是POD类型（包括原始类型和由它们构成的结构体）时，它的值成立，如果不是这样（如非平凡构造函数）。

 模板< typename T> class is_pod 
 {
 public：
 
 typedef char是; 
 typedef struct {char a [2];} No; 
 
模板< typename C> static是test（int）
 {
 union {T validPodType;} u; 
} 
 template< typename C>静态无测试（...）
 {
} 
枚举{value =（sizeof（test< T>（0））== sizeof 
}; 
 
 class NonPOD 
 {
 public：
 NonPod（const NonPod&）; 
 virtual〜NonPOD（）; 
}; 
 
 int main（）
 {
 bool a = is_pod< char> :: value; 
 bool b = is_pod< NonPOD> :: value; 
 if（a）
 printf（char is POD\\\
）; 
 if（b）
 printf（NonPOD is POD？！？！？\\\
）; 
 return 0;问题是，不仅VS 2005不具有TR1，它赢得了$不关心上面的联合（当模板参数不是POD时，它不应该有效），因此a和b的值都为true。
 
 
 
 
 
 
 
 
感谢您下面的答案。仔细阅读他们（和代码）后，我意识到，我想做的是一个错误的方法。这个想法是将SFINAE行为与模板* must_be_pod *（我在书不完美的C ++ 中找到，但是也可以在其他地方找到）的适配相结合。实际上，这将需要一个相当特殊的规则对SFINAE，这不是标准定义，显然。 
解决方案
你的方法最大的问题是你不做SFINAE这里 -  SFINAE仅适用于参数类型和返回类型。
 
 
 
但是，在标准中的所有SFINAE情境中，没有一个适用于您的情况。它们是
 
 
 
 
 
 void，引用，函数或无效大小的数组
 
 
不是类型
 
 
指向引用，引用引用，对void的引用
 
 
指向非类类型成员的指针
 
 
模板值参数的无效转换
 
 
函数类型的参数类型为void 
 
  const / volatile函数类型 
 
 
 
 
这可能是为什么在Boost文档中：
 
 
没有一些（尚未指定）从编译器帮助
，ispod永远不会
报告类或结构是一个
 POD;这总是安全的，如果可能
次优。目前（2005年5月）只有
 MWCW 9和Visual C ++ 8有
必要的编译器_intrinsics。
 
 
 
The original title here was
Workaround for SFINAE bug in VS2005 C++

This is tentative use of SFINAE to make the equivalent for the is_pod template class that exists in TR1 (In VS2005 there's no TR1 yet). It should have its value member true when the template parameter is a POD type (including primitive types and structs made of them) and false when it's not (like with non-trivial constructors). 
template <typename T> class is_pod
{
  public:

    typedef char Yes;
    typedef struct {char a[2];} No;

    template <typename C> static Yes test(int)
    {
      union {T validPodType;} u;
    }
    template <typename C> static No test(...)
    {
    }
    enum {value = (sizeof(test<T>(0)) == sizeof(Yes))};
};

class NonPOD
{
  public:
    NonPod(const NonPod &);
    virtual ~NonPOD();
};

int main()
{
  bool a = is_pod<char>::value;
  bool b = is_pod<NonPOD>::value;
  if (a) 
    printf("char is POD\n");
  if (b)
    printf("NonPOD is POD ?!?!?\n");
  return 0;
}
The problem is, not only VS 2005 doesn't have TR1, it won't care about the union above (which shouldn't be valid when the template parameter is not a POD), so both a and b evaluate to true.



Thanks for the answers posted below. After reading carefully them (and the code) I realized that what I was trying to do was really a wrong approach. The idea was to combine SFINAE behavior with an adaptation to the template *must_be_pod* (which I found in the book Imperfect C++, but it can be found in another places, too). Actually, this would require a quite particular set of rules for SFINAE, which are not what the standard defines, obviously. This is not really a bug in VS, after all.
解决方案
The biggest problem with your approach is you don't do SFINAE here - SFINAE only applies to parameter types and return type here.

However, of all the SFINAE situations in the standard, none applies to your situation. They are


arrays of void, references, functions, or of invalid size
type member that is not a type
pointers to references, references to references, references to void
pointer to member of a non-class type
invalid conversions of template value parameters
function types with arguments of type void
const/volatile function type


That's probably why in Boost documentation, there is:

  
Without some (as yet unspecified) help
  from the compiler, ispod will never
  report that a class or struct is a
  POD; this is always safe, if possibly
  sub-optimal. Currently (May 2005) only
  MWCW 9 and Visual C++ 8 have the
  necessary compiler-_intrinsics.


                        
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