成员运算符的C ++ enable_if（或解决方法）
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template<typename 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\'
}
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佩疵瓦
解决了此问题，如下所示：typename std::add_lvalue_reference<T>::type operator*() const { return *p; }珊畴炮贩号
一起使用）移动到基类中，并使make4ѭ从中继承。那么就可以专门化为not3ѭ不声明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> { };驮帽俺篮号
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> { };奥李
typename disable_if<is_void<T>, T>::type& operator* () const { return *p;}苦诫
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); } };