Variable length Struct NonStandard in C++11? [duplicate]

Answer 1

No, you cannot do it compliantly *.

Use a std::vector.

^{* I'm assuming that C++ doesn't add any rules that contradict C in this area. IMO it's highly unlikely, though I don't have time to verify that at the minute.}

Answer 2

This is C++, not C. You don't need this flexible array member hack in C++, because you can easily make a template class which can endow any struct with a flexible array past the end and encapsulate the pointer arithmetic calculation and the memory allocation to make it work. Watch:

#include <cstring>

template <typename STRUCT, typename TYPE> class flex_struct {
public:
  TYPE *tail()
  {
    return (TYPE *) ((char *) this + padded_size());
  }

  // substitute malloc/free here for new[]/delete[] if you want
  void *operator new(size_t size, size_t tail)
  {
    size_t total = padded_size() + sizeof (TYPE) * tail;
    return new char[total];
  }

  void operator delete(void *mem)
  {
    delete [] (char *) mem;
  }
private:
  static size_t padded_size() {
    size_t padded = sizeof (flex_struct<STRUCT, TYPE>);
    if(padded % alignof(TYPE) != 0) {
         padded = padded & ~(alignof(TYPE)-1) + alignof(TYPE);
    }
    return padded;
  }
};

struct mystruct : public flex_struct<mystruct, char> {
  int regular_member;
};

int main()
{
  mystruct *s = new (100) mystruct; // mystruct with 100 chars extra
  char *ptr = s->tail();            // get pointer to those 100 chars
  memset(ptr, 0, 100);              // fill them
  delete s;                         // blow off struct and 100 chars
}

Answer 3

The struct hack was never standard. This should be a standard viable replacement:

struct MyStruct {
    uint size;
    int32 buf[1];
};