[英]Using SFINAE to detect POD-ness of a type in C++
The original title here was Workaround for SFINAE bug in VS2005 C++这里的原标题是VS2005 C++ 中 SFINAE 错误的解决方法
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).这是 SFINAE 的尝试性使用,以生成 TR1 中存在的 is_pod 模板类的等效项(在 VS2005 中还没有 TR1)。 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).
当模板参数是 POD 类型(包括原始类型和由它们组成的结构)时,它的值成员应该是 true,否则它应该是 false(就像非平凡的构造函数一样)。
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.问题是,不仅 VS 2005 没有 TR1,它也不会关心上面的联合(当模板参数不是 POD 时它应该无效),所以 a 和 b 评估为真。
Thanks for the answers posted below.感谢您在下面发布的答案。 After reading carefully them (and the code) I realized that what I was trying to do was really a wrong approach.
仔细阅读它们(和代码)后,我意识到我试图做的确实是一种错误的方法。 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).
这个想法是将 SFINAE 行为与对模板must_be_pod的适应相结合(我在Imperfect C++一书中找到了它,但它也可以在其他地方找到)。 Actually, this would require a quite particular set of rules for SFINAE, which are not what the standard defines, obviously.
实际上,这将需要一套非常特殊的 SFINAE 规则,这显然不是标准定义的。 This is not really a bug in VS, after all.
毕竟,这并不是 VS 中的真正错误。
The biggest problem with your approach is you don't do SFINAE here - SFINAE only applies to parameter types and return type here.你的方法最大的问题是你不在这里做 SFINAE - SFINAE 只适用于参数类型和返回类型。
However, of all the SFINAE situations in the standard, none applies to your situation.但是,在标准中的所有 SFINAE 情况中,没有一个适用于您的情况。 They are
他们是
That's probably why in Boost documentation, there is:这可能就是为什么在 Boost 文档中,有:
Without some (as yet unspecified) help from the compiler, ispod will never report that a class or struct is a POD;
如果没有编译器的一些(尚未指定的)帮助,ispod 永远不会报告类或结构是 POD; this is always safe, if possibly sub-optimal.
这总是安全的,如果可能是次优的。 Currently (May 2005) only MWCW 9 and Visual C++ 8 have the necessary compiler-_intrinsics.
目前(2005 年 5 月)只有 MWCW 9 和 Visual C++ 8 具有必要的编译器-_intrinsics。
This doesn't work with VS2008 either, but I suspect you knew that too.这也不适用于 VS2008,但我怀疑您也知道这一点。 SFINAE is for deducing template arguments for template parameters;
SFINAE 用于为模板参数推导模板参数; you can't really deduce the type of something that reveals the constructor-ness of a type, even though you can create a type that is incompatible with another type (ie, unions can't use non-POD).
即使您可以创建与另一种类型不兼容的类型(即联合不能使用非 POD),您也无法真正推断出揭示类型构造函数的事物的类型。
In fact, VS 2008 uses compiler support for traits to implement std::tr1::type_traits
.事实上,VS 2008 使用编译器对特征的支持来实现
std::tr1::type_traits
。
I'm not sure about the way you're trying to do SFINAE here, since is_pod<T>::test(...)
will match is_pod<T>::test(0)
too.我不确定你在这里尝试做 SFINAE 的方式,因为
is_pod<T>::test(...)
也会匹配is_pod<T>::test(0)
。 Perhaps if you use a different type instead of 'int' you'd get a better match:也许如果你使用不同的类型而不是 'int' 你会得到更好的匹配:
template <typename T> class is_pod
{
struct my_special_type { };
public:
typedef char Yes;
typedef struct {char a[2];} No;
template <typename C> static Yes test(my_special_type)
{
union {T validPodType;} u;
}
template <typename C> static No test(...)
{
}
enum {value = (sizeof(test<T>(my_special_type())) == sizeof(Yes))};
};
You might also want to look at Boost.Enable_i f to do your SFINAE for you -- unless you're trying to implement your own library or for some reason.您可能还想查看Boost.Enable_if来为您执行 SFINAE —— 除非您正在尝试实现自己的库或出于某种原因。
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