检查成员存在,可能在基类C ++ 11中 [英] Checking a member exists, possibly in a base class, C++11 version
问题描述
在 http://stackoverflow.com/a/1967183/134841 中,提供了一种解决方案,用于静态检查成员存在,可能在类型的子类中:
template< typename Type>
class has_resize_method
{
class yes {char m;};
class no {yes m [2];};
struct BaseMixin
{
void resize(int){}
};
struct Base:public Type,public BaseMixin {};
template< typename T,T t> class Helper {};
template< typename U>
static no deduce(U *,Helper< void(BaseMixin :: *)(),& U :: foo> * = 0);
static yes deduce(...);
public:
static const bool result = sizeof(yes)== sizeof(deduce((Base *)(0)));
};
但是,它不适用于C ++ 11 final
类,因为它从测试类继承,
final
阻止。
<这一个:
模板< typename C>
struct has_reserve_method {
private:
struct No {};
struct Yes {No no [2]; };
template< typename T,typename I,void(T :: *)(I)> struct sfinae {};
template< typename T>静态无检查(...);
template< typename T> static是检查(sfinae< T,int,& T :: reserve> *);
template< typename T> static是检查(sfinae< T,size_t,& T :: reserve> *);
public:
static const bool value = sizeof(check< C>(0))== sizeof(Yes);
};
无法找到 reserve(int / size_t)$ c $
这个元函数的实现在基类中找到 reserved()
T
并且仍然有效,如果 T
是 final
p>
实际上,在C ++ 11中,由于 decltype
和延迟返回绑定机制。
现在,使用方法测试这个更简单:
//如果保留不存在或不可访问,则由SFINAE舍弃
template< typename T>
constexpr auto has_reserve_method(T& t) - > decltype(t.reserve(0),bool()){
return true;
}
//当SFINAE舍弃模板方法时用作回退
constexpr bool has_reserve_method(...){return false; }
然后,您可以在类中使用这个例子:
template< typename T,bool b>
struct Reserver {
static void apply(T& t,size_t n){t.reserve(n); }
};
template< typename T>
struct Reserver< T,false> {
static void apply(T& t,size_t n){}
};
您使用它:
template< typename T>
bool reserve(T& t,size_t n){
Reserver< T,has_reserve_method(t)> :: apply(t,n);
return has_reserve_method(t);
}
或者您可以选择 enable_if
方法:
template< typename T>
自动预留(T& t,size_t n) - > typename std :: enable_if< has_reserve_method(t),bool> :: type {
t.reserve(n);
return true;
}
template< typename T>
自动预留(T& t,size_t n) - > typename std :: enable_if< not have_reserve_method(t),bool> :: type {
return false;
}
请注意,这种切换事实上并不那么容易。一般来说,只要SFINAE存在就容易多了 - 你只需要 enable_if
一个方法,而不提供任何回退:
template< typename T>
自动预留(T& t,size_t n) - > decltype(t.reserve(n),void()){
t.reserve(n);
}
如果替换失败,则从可能的重载列表中删除此方法。 / p>
注意:由于,
(逗号运算符)的语义, code> decltype ,只有最后一个实际决定了类型。方便检查多个操作。
In http://stackoverflow.com/a/1967183/134841, a solution is provided for statically checking whether a member exists, possibly in a subclass of a type:
template <typename Type>
class has_resize_method
{
class yes { char m;};
class no { yes m[2];};
struct BaseMixin
{
void resize(int){}
};
struct Base : public Type, public BaseMixin {};
template <typename T, T t> class Helper{};
template <typename U>
static no deduce(U*, Helper<void (BaseMixin::*)(), &U::foo>* = 0);
static yes deduce(...);
public:
static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));
};
However, it doesn't work on C++11 final
classes, because it inherits from the class under test, which final
prevents.
OTOH, this one:
template <typename C>
struct has_reserve_method {
private:
struct No {};
struct Yes { No no[2]; };
template <typename T, typename I, void(T::*)(I) > struct sfinae {};
template <typename T> static No check( ... );
template <typename T> static Yes check( sfinae<T,int, &T::reserve> * );
template <typename T> static Yes check( sfinae<T,size_t,&T::reserve> * );
public:
static const bool value = sizeof( check<C>(0) ) == sizeof( Yes ) ;
};
fails to find the reserve(int/size_t)
method in baseclasses.
Is there an implementation of this metafunction that both finds reserved()
in baseclasses of T
and still works if T
is final
?
Actually, things got much easier in C++11 thanks to the decltype
and late return bindings machinery.
Now, it's just simpler to use methods to test this:
// Culled by SFINAE if reserve does not exist or is not accessible
template <typename T>
constexpr auto has_reserve_method(T& t) -> decltype(t.reserve(0), bool()) {
return true;
}
// Used as fallback when SFINAE culls the template method
constexpr bool has_reserve_method(...) { return false; }
You can then use this in a class for example:
template <typename T, bool b>
struct Reserver {
static void apply(T& t, size_t n) { t.reserve(n); }
};
template <typename T>
struct Reserver <T, false> {
static void apply(T& t, size_t n) {}
};
And you use it so:
template <typename T>
bool reserve(T& t, size_t n) {
Reserver<T, has_reserve_method(t)>::apply(t, n);
return has_reserve_method(t);
}
Or you can choose a enable_if
method:
template <typename T>
auto reserve(T& t, size_t n) -> typename std::enable_if<has_reserve_method(t), bool>::type {
t.reserve(n);
return true;
}
template <typename T>
auto reserve(T& t, size_t n) -> typename std::enable_if<not has_reserve_method(t), bool>::type {
return false;
}
Note that this switching things is actually not so easy. In general, it's much easier when just SFINAE exist -- and you just want to enable_if
one method and not provide any fallback:
template <typename T>
auto reserve(T& t, size_t n) -> decltype(t.reserve(n), void()) {
t.reserve(n);
}
If substitution fails, this method is removed from the list of possible overloads.
Note: thanks to the semantics of ,
(the comma operator) you can chain multiple expressions in decltype
and only the last actually decides the type. Handy to check multiple operations.
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