It's easy enough to make the QVector
work for a non-default-constructible type T:
#define QVECTOR_NON_DEFAULT_CONSTRUCTIBLE(Type)
template <> QVector<Type>::QVector(int) = delete;
template <> void QVector<Type>::resize(int newSize) {
Q_ASSERT(newSize <= size());
detach();
}
template <> void QVector<Type>::defaultConstruct(Type*, Type*) { Q_ASSERT(false); }
The macro needs to be present right after MyType
declaration - in the header file (if any), and it must be in namespace or global scope:
struct MyType { ... };
QVECTOR_NON_DEFAULT_CONSTRUCTIBLE(MyType)
struct A {
struct MyType2 { ... };
};
QVECTOR_NON_DEFAULT_CONSTRUCTIBLE(A::MyType2);
No, the wrapper is not correct. It doesn't destruct the object
member. It also doesn't offer move semantics, doesn't protect from being default-constructed, etc. The hack
union member is not necessary. Nothing in a union will be default-constructed for you.
Here's a more correct wrapper - it pretty much resembles std::optional
. See here to see how much nuance an optional
needs :)
// https://github.com/KubaO/stackoverflown/tree/master/questions/vector-nodefault-33380402
template <typename T> class Wrapper final {
union {
T object;
};
bool no_object = false;
void cond_destruct() {
if (!no_object)
object.~T();
no_object = true;
}
public:
Wrapper() : no_object(true) {}
Wrapper(const Wrapper &o) : no_object(o.no_object) {
if (!no_object)
new (&object) T(o.object);
}
Wrapper(Wrapper &&o) : no_object(o.no_object) {
if (!no_object)
new (&object) T(std::move(o.object));
}
Wrapper(const T &o) : object(o) {}
Wrapper(T &&o) : object(std::move(o)) {}
template <class...Args> Wrapper(Args...args) : object(std::forward<Args>(args)...) {}
template <class U, class...Args> Wrapper(std::initializer_list<U> init, Args...args) :
object(init, std::forward<Args>(args)...) {}
operator T& () & { assert(!no_object); return object; }
operator T&& () && { assert(!no_object); return std::move(object); }
operator T const&() const& { assert(!no_object); return object; }
Wrapper &operator=(const Wrapper &o) & {
if (no_object)
::new (&object) T(o);
else
object = o.object;
no_object = false;
return *this;
}
Wrapper &operator=(Wrapper &&o) & {
if (no_object)
::new (&object) T(std::move(o.object));
else
object = std::move(o.object);
no_object = false;
return *this;
}
template<class... Args> T &emplace(Args&&... args) {
cond_destruct();
::new (&object) T(std::forward<Args>(args)...);
no_object = false;
return object;
}
~Wrapper() {
cond_destruct();
}
};
Since the assignment operators are ref-qualified, it disallows assigning to rvalues, so it has the IMHO positive property that the following won't compile:
Wrapper<int>() = 1 // likely Wrapper<int>() == 1 was intended
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