How to initialize a sequence of non-movable, non-copyable objects? - c++

Question

Let's say I have a type which is neither movable nor copyable:
struct foo
{
explicit foo( size_t ){}
~foo(){}
foo( foo const & ) = delete;
foo( foo && ) = delete;
foo& operator=( foo const & ) = delete;
foo& operator=( foo & ) = delete;
};
Now given a number known at compile time (call it N), is there any way that I can create a "sequence" of these on the stack with each one initialized with numbers 0 through N-1? I would be satisfied with a C-style array foo[N], a std::array< foo, N >, or perhaps even a std::tuple of some kind.
What I'm trying to avoid is writing out:
foo f0( 0 ), f1( 1 ), ... fNminus1( N-1 );
when it feels like this is something the compiler should be able to do for me. The best I've been able to come up with is using boost::optional.
boost::optional< foo > f[N];
for( size_t i = 0U; i < N; ++i )
f[i] = boost::in_place( i );
But that relies on runtime logic even though all the required information is available at compile-time. Plus, I'm left with something that behaves like an array of pointers.

Answer 1

// create a type with the proper alignment
typedef std::aligned_storage<sizeof(foo), std::alignment_of<foo>::value>::type buffer_type;
const int N = 10;
// create an array of uninitialized raw data
buffer_type storage_buffer[N];
// initialize each foo object with placement new
for (size_t i=0; i<N; ++i)
new (storage_buffer + i) foo(i);
foo * fp = (foo*)(&storage_buffer);
// access your foo objects via fp
// you must manually call the destructor of each object
for (size_t i=0; i<N; ++i)
fp[i].~foo();
If that seems like a lot of hassle, it is. But you could easily encapsulate that functionality in a class.

Answer 2

Although not strictly an array, you can sort of accomplish this with template recursion
template< typename T, size_t N >
struct type_array : public type_array< T, N-1 > {
// this is the Nth element
T elem;
// it is constructed with N
type_array() : elem( N ) {}
// member function to return the Nth element
T & get( size_t n ) {
if ( n == N ) {
return elem;
} else {
return type_array< T, N-1 >::get( n );
}
}
};
// base case when N == 0
template< typename T >
struct type_array<T, 0> {
T elem;
type_array() : elem( 0 ) {}
T & get( size_t n ) {
return elem;
}
};
Usage:
type_array< foo, 100 > foo_array; // construct 100 foos
foo_array.get(1); // foo with n == 1
foo_array.get(2); // foo with n == 2

Answer 3

Like the answer from Benjamin Lindley, but packed in a class:
#include <type_traits>
#include <utility>
#include <new>
template<typename T>
class uninitialized {
public:
constexpr uninitialized() { }
~uninitialized() {
get().~T();
}
explicit uninitialized(const uninitialized& other) {
construct(other);
}
explicit uninitialized(uninitialized&& other) {
construct(std::move(other));
}
template<class... Args>
explicit uninitialized(Args&&... args) {
construct(std::forward<Args>(args)...);
}
template<class... Args>
void construct(Args&&... args) noexcept {
static_assert(std::is_nothrow_constructible<T, Args...>::value, "constructor should not throw!");
::new(getPointer()) T (std::forward<Args>(args)...);
}
uninitialized& operator = (const T& t) {
get() = t;
return *this;
}
uninitialized& operator = (T&& t) {
get() = std::move(t);
return *this;
}
T* operator -> () { return getPointer(); }
T& operator * () { return get(); }
T* operator & () { return getPointer(); }
T* getPointer() { return reinterpret_cast<T*>(&data); }
T& get() { return *reinterpret_cast<T*>(&data); }
const T* operator -> () const { return getPointer(); }
const T& operator * () const { return get(); }
const T* operator & () const { return getPointer(); }
const T* getPointer() const { return reinterpret_cast<const T*>(&data); }
const T& get() const { return *reinterpret_cast<const T*>(&data); }
private:
std::aligned_storage<sizeof(T), std::alignment_of<T>::value>::type data;
};
Now things are a little bit easier:
uninitialized<foo> f[N];
for (size_t i = 0; i < N; ++i)
f[i].construct(i);
for (const auto& fooref : f)
fooref->bar();
// foo::~foo is called for you