如何解决如果一个不知道其参数类型的人怎么能默认一个特殊的成员函数？
| 考虑这种情况：template<typename T>
struct A {
A(A ???&) = default;
A(A&&) { /* ... */ }
T t;
};
A(A const&) = default; // or
A(A &) = default; // ?
A &
A &
A const&
A &
A const&
A &
template<typename T>
struct A {
A(A &);
A(A const&);
A(A&&) { /* ... */ }
T t;
};
// valid!?
template<typename T> A<T>::A(A const&) = default;
template<typename T> A<T>::A(A &) = default;
A const&
A &
解决方法
在我看来，您将需要某种类型的演绎（概念？）。
除非成员之一要求将其写为“ 15”，否则编译器通常将使用“ 14”版本。因此，我们可以包装一些小模板黑客，以检查每个成员具有哪个版本的副本构造函数。
最新
请在ideone上进行咨询，或在代码段之后按Clang阅读错误。
#include <memory>
#include <type_traits>
template <bool Value,typename C>
struct CopyConstructorImpl { typedef C const& type; };
template <typename C>
struct CopyConstructorImpl<false,C> { typedef C& type; };
template <typename C,typename T>
struct CopyConstructor {
typedef typename CopyConstructorImpl<std::is_constructible<T,T const&>::value,C>::type type;
};
// Usage
template <typename T>
struct Copyable {
typedef typename CopyConstructor<Copyable<T>,T>::type CopyType;
Copyable(): t() {}
Copyable(CopyType) = default;
T t;
};
int main() {
{
typedef Copyable<std::auto_ptr<int>> C;
C a; C const b;
C c(a); (void)c;
C d(b); (void)d; // 32
}
{
typedef Copyable<int> C;
C a; C const b;
C c(a); (void)c;
C d(b); (void)d;
}
}
6167745.cpp:32:11: error: no matching constructor for initialization of \'C\' (aka \'Copyable<std::auto_ptr<int> >\')
C d(b); (void)d;
^ ~
6167745.cpp:22:7: note: candidate constructor not viable: 1st argument (\'const C\' (aka \'const Copyable<std::auto_ptr<int> >\')) would lose const qualifier
Copyable(CopyType) = default;
^
6167745.cpp:20:7: note: candidate constructor not viable: requires 0 arguments,but 1 was provided
Copyable(): t() {}
^
1 error generated.
#include <memory>
#include <type_traits>
// Usage
template <typename T>
struct Copyable
{
static bool constexpr CopyByConstRef = std::is_constructible<T,T const&>::value;
static bool constexpr CopyByRef = !CopyByConstRef && std::is_constructible<T,T&>::value;
Copyable(): t() {}
Copyable(Copyable& rhs,typename std::enable_if<CopyByRef>::type* = 0): t(rhs.t) {}
Copyable(Copyable const& rhs,typename std::enable_if<CopyByConstRef>::type* = 0): t(rhs.t) {}
T t;
};
int main() {
{
typedef Copyable<std::auto_ptr<int>> C; // 21
C a; C const b; // 22
C c(a); (void)c; // 23
C d(b); (void)d; // 24
}
{
typedef Copyable<int> C; // 27
C a; C const b; // 28
C c(a); (void)c; // 29
C d(b); (void)d; // 30
}
}
6167745.cpp:14:78: error: no type named \'type\' in \'std::enable_if<false,void>\'
Copyable(Copyable const& rhs,typename std::enable_if<CopyByConstRef>::type* = 0): t(rhs.t) {}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
6167745.cpp:22:11: note: in instantiation of template class \'Copyable<std::auto_ptr<int> >\' requested here
C a; C const b;
^
6167745.cpp:13:67: error: no type named \'type\' in \'std::enable_if<false,void>\'
Copyable(Copyable& rhs,typename std::enable_if<CopyByRef>::type* = 0): t(rhs.t) {}
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
6167745.cpp:28:11: note: in instantiation of template class \'Copyable<int>\' requested here
C a; C const b;
^
6167745.cpp:24:11: error: no matching constructor for initialization of \'C\' (aka \'Copyable<std::auto_ptr<int> >\')
C d(b); (void)d;
^ ~
6167745.cpp:13:7: note: candidate constructor not viable: 1st argument (\'const C\' (aka \'const Copyable<std::auto_ptr<int> >\')) would lose const qualifier
Copyable(Copyable& rhs,typename std::enable_if<CopyByRef>::type* = 0): t(rhs.t) {}
^
6167745.cpp:11:7: note: candidate constructor not viable: requires 0 arguments,but 1 was provided
Copyable(): t() {}
^
std::auto_ptr
A&
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