placement new + array +alignment

Question

SomeObj<unsigned int>* Buffer;
char* BufferPtr = MemoryManager::giveMeSomeBytes(resX*resY*sizeof(SomeObj<unsigned int>));
Buffer = new(BufferPtr) SomeObj<unsigned int>[resX*resY];

when I step past these lines with the debugger, it shows me the values for the variables Buffer and BufferPtr:

BufferPtr: 0x0d7f004c
Buffer:    0x0d7f0050

I don't really understand why those values differ. The way I understand it, placement new should use the memory starting at address 'BufferPtr' to initialize the array elements using theyr default constructors on the allocated memory and return a pointer to the first byte of the first element in the array, which should be exactly the same byte as passed to the placement new operator.

Did I understand something wrong or can someone tell me why the values differ?

thanks!

//edit: ok - i investigated the issue further and got more confusing results:

    int size = sizeof(matth_ptr<int>);

    char* testPtr1 = (char*)malloc(a_resX*a_resY*sizeof(int));
    int* test1 = new(testPtr1) int[a_resX*a_resY];

    char* testPtr2 = mmgr::requestMemory(a_resX*a_resY*sizeof(int));
    int* test2 = new(testPtr2) int[a_resX*a_resY];

    char* testPtr3 = (char*)malloc(a_resX*a_resY*sizeof(matth_ptr<int>));
    matth_ptr<int>* test3 = new(testPtr3)matth_ptr<int>[a_resX*a_resY];

    char* testPtr4 = mmgr::requestMemory(a_resX*a_resY*sizeof(matth_ptr<int>));
    matth_ptr<int>* test4 = new(testPtr4)matth_ptr<int>[a_resX*a_resY];

the debugger returns me the following values for my variables:

size: 4

testPtr1:0x05100418
test1:   0x05100418
testPtr2:0x0da80050
test2:   0x0da80050

testPtr3:0x05101458
test3:   0x0510145c
testPtr4:0x0da81050
test4:   0x0da81054

so it clearly must have something to do with my generic smartpointer class matth_ptr so here it is:

template <class X> class matth_ptr
{
public:
    typedef X element_type;

    matth_ptr(){
        memoryOfst = 0xFFFFFFFF;
    } 

    matth_ptr(X* p) 
    {
        unsigned char idx = mmgr::getCurrentChunkIdx();
        memoryOfst = (int)p-(int)mmgr::getBaseAddress(idx);
        assert(memoryOfst<=0x00FFFFFF || p==0);//NULL pointer is not yet handled
        chunkIdx = idx;
    }
    ~matth_ptr()                {}
    X& operator*()              {return *((X*)(mmgr::getBaseAddress(chunkIdx)+(memoryOfst&0x00FFFFFF)));}
    X* operator->()             {return  ((X*)(mmgr::getBaseAddress(chunkIdx)+(memoryOfst&0x00FFFFFF)));}
    X* get()                    {return  ((X*)(mmgr::getBaseAddress(chunkIdx)+(memoryOfst&0x00FFFFFF)));}


    template<typename T>
    matth_ptr(const matth_ptr<T>& other) {memoryOfst=other.memoryOfst;}//put these two operators into the private part in order to prevent copying of the smartpointers
    template<typename T>
    matth_ptr& operator=(const matth_ptr<T>& other) {memoryOfst = other.memoryOfst; return *this;}
    template<typename T>
    friend class matth_ptr;
private:

    union //4GB adressable in chunks of 16 MB
    {
        struct{
            unsigned char padding[3]; //3 bytes padding
            unsigned char chunkIdx; //8 bit chunk index
        };
        unsigned int memoryOfst; //24bit address ofst
    };

};

can anyone explain me what's going on? thanks!

Answer 1

Be careful with placement new on arrays. In the current standard look to section 5.3.4.12, you'll find this:

new(2,f) T[5] results in a call of operator new[](sizeof(T)*5+y,2,f)

It is clear that it will expect the placement new operator to allocate it additional space beyond what the array contents need. "y" is specified only as a non-negative integral value. It will then offset the result of the new function by this amount.

Also look to 18.4.1.3.4 where it says the placement new operator simply returns the provided pointer. This is obviously the expected part.

Based on 5.3.4.12, since that offset may be different for every invocation of the array, the standard basically means there is no way to allocate the exact amount of size needed. In practice that value is probably constant and you could just add it to the allocation, but his amount may change per platform, and again, per invocation as the standard says.

Answer 2

您正在使用new运算符的数组版本，在您的实现中，该运算符在内存分配的前几个字节中存储有关数组大小的信息。

Answer 3

@Mat, This is actually a great question. When I've used placement new[], I've had trouble deleting the storage. Even if I call my own symmetrical placement delete[], the pointer address is not the same as was returned by my own placement new[]. This makes placement new[] completely useless, as you've suggested in the comments.

The only solution I've found was suggested by Jonathan@: Instead of placement new[], use placement new (non-array) on each of the elements of the array. This is fine for me as I store the size myself. The problem is that I have to worry about pointer alignments for elements, which new[] is supposed to do for me.

Answer 4

As others have said, this is due to your C++ implementation storing the size of the array at the start of the buffer you pass to array placement new.

An easy fix for this is to simply assign your array pointer to the buffer, then loop over the array and use regular (non-array) placement new to construct each object in the buffer.