Reputation: 7799
I deleted a copy constructor in base class but I can't get if the compiler will create an implicit copy constructor in child classes? Or does a deleted constructor in the base class prevent it?
template <typename val_t>
class exp_t {
public:
using vals_t = std::vector<val_t>;
exp_t() {}
exp_t(const exp_t<val_t> &) = delete;
exp_t(exp_t &&) = default;
virtual ~exp_t() {}
exp_t<val_t> &operator=(const exp_t<val_t> &) = delete;
exp_t<val_t> &operator=(exp_t<val_t> &) = delete;
exp_t<val_t> &operator=(exp_t &&) = default;
};
template <typename val_t>
class fact_t: public exp_t<val_t> {
using vals_t = std::vector<val_t>;
val_t m_value;
public:
fact_t(val_t &&value) : m_value{std::forward<val_t>(value)} {}
fact_t(fact_t &&) = default;
};
Will fact_t
have an implicit copy constructor? (GCC 7)
Upvotes: 6
Views: 1928
Reputation: 8333
No, as the default copy constructor would call the parent's copy constructor (which is deleted), this won't work.
Why didn't you simply test it:
int main() {
auto x = fact_t<int>(5);
auto y = x;
}
Result:
copytest.cpp: In function 'int main()':
copytest.cpp:32:14: error: use of deleted function 'fact_t<int>::fact_t(const fact_t<int>&)'
auto y = x;
^
copytest.cpp:21:7: note: 'fact_t<int>::fact_t(const fact_t<int>&)' is implicitly declared as deleted because 'fact_t<int>' declares a move constructor or move assignment operator
class fact_t: public exp_t<val_t> {
^~~~~~
Upvotes: 6