CODEWITHSUNDEEP
c++
merlin2011
merlin2011

Reputation: 75649

Is there a c++ counterpart to posix_memalign?

When I call posix_memalign to allocate aligned memory for an object of type Foo in my C++ code, I am required to do a reinterpret_cast of the address of that pointer to void**.

In general when I encounter this situation, it implies that I am missing some language feature. That is, it feels like I am calling malloc in c++ when I should be calling new. , Is there a type-aware new equivalent for aligned memory allocation in c++?

Upvotes: 6

Views: 4719

Answers (3)

nneonneo
nneonneo

Reputation: 179717

C++11 has added native language support for alignment declarations and aligned allocation.

You can specify alignas(N) on your type in C++11 to specify the minimum alignment for new objects, which the default new will respect.

Example from cppreference:

struct alignas(16) sse_t { // SSE-safe struct aligned on 16-byte boundaries
    float v[4];
};

then you can simply do

sse_t *ssevec = new sse_t;

For a replacement for posix_memalign, you can use std::aligned_storage<sizeof(T), N>, also in C++11.

Upvotes: 1

Yakk - Adam Nevraumont
Yakk - Adam Nevraumont

Reputation: 275966

I will start with the core advice first.

Foo* aligned_foo() {
  void* raw = 0;
  if(posix_memalign(&raw, 8, sizeof(Foo)))
    return 0; // we could throw or somehow communicate the failure instead
  try{
    return new(raw) Foo();
  }catch(...){
    free(raw);
    throw;
  }
}

then when you are done with the Foo* foo, do a foo->~Foo(); free(foo); instead of delete.

Note the lack of reinterpret_casts.

Here is an attempt to make it generic:

// note: stateless.  Deleting a derived with a base without virtual ~base a bad idea:
template<class T>
struct free_then_delete {
  void operator()(T*t)const{
    if(!t)return;
    t->~T();
    free(t);
  };
};
template<class T>
using aligned_ptr=std::unique_ptr<T,free_then_delete<T>>;

// the raw version.  Dangerous, because the `T*` requires special deletion:
template<class T,class...Args>
T* make_aligned_raw_ptr(size_t alignment, Args&&...args) {
  void* raw = 0;
  if(int err = posix_memalign(&raw, alignment, sizeof(T)))
  {
    if (err==ENOMEM)
      throw std::bad_alloc{};
    return 0; // other possibility is bad alignment: not an exception, just an error
  }
  try {
    // returns a T*
    return new(raw) T(std::forward<Args>(args)...);
  } catch(...) { // the constructor threw, so clean up the memory:
    free(raw);
    throw;
  }
}
template<class T,class...Args> // ,class... Args optional
aligned_ptr<T> make_aligned_ptr(size_t alignment=8, Args&&...args){
  T* t = make_aligned_raw_ptr<T>(alignment, std::forward<Args>(args)...);
  if (t)
    return aligned_ptr<T>(t);
  else
    return nullptr;
}

The unique_ptr alias aligned_ptr bundles the destroyer along with the pointer -- as this data requires destruction and free, not delete, this makes it clear. You can still .release() the pointer out, but you still have to do the steps.

Upvotes: 5

Jay Miller
Jay Miller

Reputation: 2234

Actually, you don't want to do a reinterpret_cast because then your Foo constructor isn't called. If you need to allocate memory from a special place, you then call placement new to construct the object in that memory:

void* alloc;
posix_memalign(&alloc, 8, sizeof(Foo));
Foo* foo = new (foo) Foo();

The only other way (pre C++11) would be overriding the new operator for your class. That works if you have a particular class that always requires this special allocation:

class Foo {
    void* operator new(size_t size) {
        void* newobj;
        posix_memalign(&newobj, 8, sizeof(Foo));
        return newobj;
    }
};

Then anytime you call new Foo() it will invoke this allocator. See http://en.cppreference.com/w/cpp/memory/new/operator_new for more information. Overriding operator new and operator delete can be done for individual classes or globally.

Upvotes: 3

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