compliant variable length struct in C++

Question

In standard C you can end a struct with an array of size 0 and then over allocate it to add a variable length dimension to the array:

struct var
{
    int a;
    int b[];
}

struct var * x=malloc(sizeof(var+27*sizeof(int)));

How can you do that in C++ in a standard (portable) way? It is okay to have a constraint of max posible size and obviously doesn't have to work on the stack

I was thinking of:

class var
{
...
private:
  int a;
  int b[MAX];
};

and then use allocators or overload new/delete to under allocate based on the required size:

(sizeof(var) - (MAX-27)*sizeof(int)

But, while it seems to work, its not something I'd want to have to maintain.

Is there a cleaner way that is fully standard/portable?

Answer 1

What's wrong with simply doing a variant of the C way?

If the structure has to remain purely POD, the C way is fine.

struct var
{
    int a;
    int b[1];

    static std::shared_ptr<var> make_var(int num_b) {
        const extra_bytes = (num_b ? num_b-1 : 0)*sizeof(int);
        return std::shared_ptr<var>(
                new char[sizeof(var)+extra_bytes ],
                [](var* p){delete[]((char*)(p));});
}

since it's a POD, everything works just like it did in C.

---

If b is not guaranteed to be POD, then things get more interesting. I haven't tested any of it, but it would look more or less like so. Note that make_var relies on make_unique, because it uses a lambda destructor. You can make it work without this, but it's more code. This works just like the C way, except it cleanly handles variable amounts of types with constructors and destructors, and handles exceptions

template<class T>
struct var {
    int a;

    T& get_b(int index) {return *ptr(index);}
    const T& get_b(int index) const {return *ptr(index);}

    static std::shared_ptr<var> make_var(int num_b);
private:
    T* ptr(int index) {return static_cast<T*>(static_cast<void*>(&b))+i;}
    var(int l);
    ~var();
    var(const var&) = delete;
    var& operator=(const var&) = delete;

    typedef typename std::aligned_storage<sizeof(T), std::alignof(T)>::type buffer_type;
    int len;
    buffer_type b[1];
};
template<class T> var::var(int l)
    :len(0)
{
    try {
        for (len=0; len<l; ++len)
            new(ptr(i))T();
    }catch(...) {
        for (--len ; len>=0; --len)
            ptr(i)->~T();
        throw;
    }
}
template<class T> var::~var()
{
    for ( ; len>=0; --len)
        ptr(i)->~T();
}
template<class T> std::shared_ptr<var> var::make_var(int num_b)
{
    const extra_bytes = (num_b ? num_b-1 : 0)*sizeof(int);
    auto buffer = std::make_unique(new char[sizeof(var)+extra_bytes ]);
    auto ptr = std::make_unique(new(&*buffer)var(num_b), [](var*p){p->~var();});
    std::shared_ptr<var> r(ptr.get(), [](var* p){p->~var(); delete[]((char*)(p));});
    ptr.release();
    buffer.release;
    return std::move(r);
}

Since this is untested, it probably doesn't even compile, and probably has bugs. I'd normally use std::unique_ptr but I'm too lazy to make proper standalone deleters, and unique_ptr is hard to return from a function when the deleter is a lambda. On the off chance you want to use code like this, use a proper standalone deleter.

Answer 2

While this is not directly answering your question -- I would point to that a better practice in C++ is to use the STL lib for this sort of variable length array -- it is safe and simpler and understood by anybody who will maintain it after you.

class var
{
...
private:
  int a;
  std::vector<int> b; // or use std::deque if more to your liking
};

Now you can just new it up like any other class;

var* myvar = new var;

And you can use it just like a old type array without explicitly allocating the memory (although that is not what most ++ programmers do)

myvar->b[0] = 123;
myvar->b[1] = 123;
myvar->b[2] = 123;