C++ enable_if (or workaround) for member operator
template<typen开发者_运维知识库ame T>
struct foo
{
T* p;
foo(T* x) : p(x) {}
~foo() { if(p) delete p; }
T& operator*() const { return *p; }
};
int main()
{
foo<int> i(new int);
foo<void> v(new int); // <= illegal use of type 'void'
}
If T = void then I don't want to implement the operator*(). How can I achieve this? I don't want to specialize the class, because there are many other methods in my class.
PS: Please note that this is just an example to explain my issue.
You can move all the other methods (which play well with T==void
) into a base class and make foo
derive from it. Then foo
can be specialized to not declare the operator*
for T==void
template <typename T>
struct foobase {
T* p;
foobase(T* x) : p(x) {}
~foobase() { if(p) delete p; }
};
template <typename T>
struct foo : foobase<T> {
T& operator*() const { return *p; }
};
template<>
struct foo<void> : foobase<void> {
};
The C++11 standard solved this for std::unique_ptr
like so:
typename std::add_lvalue_reference<T>::type
operator*() const { return *p; }
Like that:
template<typename T>
struct foo_base
{
T* p;
foo(T* x) : p(x) {}
~foo() { if(p) delete p; }
// other methods …
};
template<typename T>
struct foo : foo_base<T>
{
T& operator*() const { return *p; }
};
template<>
struct foo<void> : foo_base<void>
{
};
How about
typename disable_if<is_void<T>, T>::type& operator* () const { return *p;}
or am I missing something obvious here?
What's about this solution:
template<typename T>
struct foo
{
T* p;
foo(T* x) : p(x) {}
~foo() { if(p) delete p; }
template<typename U> struct impl { U& deref(U* p) { return *p; } };
template<> struct impl<void> { void deref(void* p) { } };
typename boost::conditional<std::is_void<T>::value, T, typename std::add_reference<T>::type>::type operator*() const
{
static_assert(!std::is_void<T>::value, "illegal use of type 'void'");
return impl<T>().deref(p);
}
};
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