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Simple storage class in C++ and strict aliasing

开发者 https://www.devze.com 2023-01-25 01:58 出处:网络
I have the following code for having a small class for storage. #include <iostream> template<typename T>

I have the following code for having a small class for storage.

#include <iostream>

template<typename T>
class storage
{
private:
  struct destroy
  {
    T& m_t;
    destroy(T& t) : m_t(t) { }
    ~destroy() { m_t.~T(); }
  };

  char m_c[sizeof(T)];
  void* address() { return &m_c[0]; }

public:
  void set(const T& t) { new (address()) T(t); }

  T get()
  {
    T& t = *static_cast<T*>(address());
    destroy _d(t);
    return t;
  }

};

template<typename T>
class choosable_storage
{
private:
  union
  {
    T*         m_p;
    storage<T> m_storage;
  };
  bool m_direct;

public:
  choosable_storage() : m_direct(false) { }

  void set_direct(const T& t)
  {
    m_direct = true;
    m_storage.set(t);
  }

  void set_indirect(T* const t) { m_p = t; }

  T get()
  {
    if (m_direct) return m_storage.get();
    return *m_p;
  }

};

int main(void)
{
  storage<int> s; // no problem开发者_高级运维s
  s.set(42);
  std::cout << s.get() << std::endl;

  int i = 10;

  choosable_storage<int> c1; // strict aliasing warnings
  c1.set_indirect(&i);
  std::cout << c1.get() << std::endl;

  choosable_storage<int> c2;
  c2.set_direct(i);
  std::cout << c2.get() << std::endl;

  return 0;
}

gcc 4.4 warns that I break the strict aliasing rules in storage::get() when I return.

AFAIK, I do not violate any rules. Do I actually violate strict aliasing or is gcc getting picky here?

And is there a way of having it warning free without disabling strict aliasing?

Thanks

EDIT:

On the other hand, the following implementation does not give any warnings:

template<typename T>
class storage
{
private:
  struct destroy
  {
    T& m_t;
    destroy(T& t) : m_t(t) { }
    ~destroy() { m_t.~T(); }
    T const& operator()() const { return m_t; }
  };

  char m_c[sizeof(T)];

public:
  void set(const T& t) { new(static_cast<void*>(m_c)) T(t); }

  T get(void) { return destroy(*static_cast<T*>(static_cast<void*>(m_c)))(); }

};

EDIT:

gcc 4.5 and up does not issue a warning - so apparently this was just a misinterpretation of the strict aliasing rules or a bug in gcc 4.4.x


Do I actually violate strict aliasing or is gcc getting picky here?

There are two different interpretation of the strict aliasing rule:

  • the usual weak strict aliasing rule: (except for char/unsigned char type) you cannot use a cast or union to perform type punning, you need memcpy (or two volatile accesses); such code really is not very reasonable, and is explicitly forbidden in C (except the latest, very absurd, C standard) and C++.
  • the unusual strong strict aliasing rule: you cannot reuse memory: once a region of memory has a "dynamic type" (a what?), it cannot be used with a different type. So you just cannot write an allocator function (malloc alternative) C, unless it only calls malloc/free each time.

The strong rule is ill-defined, break lots of reasonable code, and for some reason was chosen by the GCC maintainers (based entirely on self delusion and circular justifications - this is really ugly). To make C++ code work with the strong strict aliasing optimisation, the maintainers of G++ added pessimisations for typical C++ code (based on more self delusion), to keep the optimisation!

I do not know if/when they realise their mistake, in case of doubt just disable strict aliasing. It is a very minor optimisation anyway.


For the purposes of this question, strict aliasing rule states, essentially, that you are not supposed to access an object except through a pointer/reference of its own type or a pointer/reference to a character type (char or unsigned char).

In your code, you have an array m_c of elements of type char, and you're trying to access it through a reference of type T. That's a strict aliasing violation. On some more exotic platforms, this could have repercussions, for example, if m_c is not properly aligned to hold the element of type T.

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