C ++模板:如何确定类型是否适合子类化 [英] C++ templates: how to determine if a type is suitable for subclassing
问题描述
假设我有一些模板类,取决于 T
类型。 T
几乎可以是任何东西: int
, int *
pair< int,int>
或 struct lol
它不能 void
,一个引用或任何cv限定。对于一些优化我需要知道,如果我可以子类 T
。所以,我需要一些特征类型 is_subclassable
,确定为基本特征的逻辑组合或通过一些SFINAE技巧。
Let's say I have some templated class depending on type T
. T
could be almost anything: int
, int*
, pair <int, int>
or struct lol
; it cannot be void
, a reference or anything cv-qualified though. For some optimization I need to know if I can subclass T
. So, I'd need some trait type is_subclassable
, determined as a logical combination of basic traits or through some SFINAE tricks.
在原始示例中, int
和 int *
不可子类化,而 pair < int,int>
和 struct lol
。
In the original example, int
and int*
are not subclassable, while pair <int, int>
and struct lol
are.
:下面指出,联合也不是子类, T
也可以是联合类型。
EDIT: As litb pointed out below, unions are also not subclassable and T
can be a union type as well.
如何写我需要的trait类型?
How do I write the trait type I need?
推荐答案
类。没有办法我知道这样做(和boost也没有找到一种方式)。如果你可以使用联合案例假阳性,你可以使用 is_class
。
You want to determine whether it is a non-union class. There is no way known to me to do that (and boost hasn't found a way either). If you can live with union cases false positives, you can use a is_class
.
template<typename> struct void_ { typedef void type; };
template<typename T, typename = void>
struct is_class { static bool const value = false; };
template<typename T>
struct is_class<T, typename void_<int T::*>::type> {
static bool const value = true;
};
Boost有一个 is_union
,它将使用编译器特定的内置,这将帮助您这里。 is_class
(其中boost也提供)结合 is_union
将解决您的问题。
Boost has an is_union
that uses compiler-specific builtins though, which will help you here. is_class
(which boost also provides) combined with is_union
will solve your problem.
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