内存对齐

void* AlignedMemory::allocate(size_t size, size_t alignment)     {         assert(0 < alignment && alignment <= 128 && Bitwise::isPO2(alignment));      unsigned char* p = new unsigned char[size + alignment];         size_t offset = alignment - (size_t(p) & (alignment-1));      unsigned char* result = p + offset;          result[-1] = (unsigned char)offset;     return result;     }

该函数是这样的，因为要对内存按2的n次方对齐，在unsigned char* p = new unsigned char[size + alignment]; 这一句先申请了内存，但所得到的指针p可能不是内存对齐的，如按4字节对齐的话，内存地址只能是0x00000000,0x00000004,0x00000008等，但如果分配到的p是0x00000001这个地址时怎么办，这时要通过加一个偏移值来使地址对齐，偏移值是这样算的，size_t offset = alignment - (size_t(p) & (alignment-1));其中( (size_t(p) & (alignment-1))等价于把p的值对alignment求余，这就是size_t offset = alignment - size_t(p) %(alignment);对0x00000001这个地址来说：offset = 4 - (0x00000001 % 4) = 3; 0x00000001 + 3 = 0x00000004，这样内存地址就对齐了，至于result[-1] = (unsigned char)offset;因为1<=OFFSET<=3所以可以空一个字节也就是一个char出来存储result 最后的一行代码将result[-1]存放offset值的作用，是为了在释放时正确算回p的位置.    void AlignedMemory::deallocate(void* p)    {        if (p)        {            unsigned char* mem = (unsigned char*)p;            mem = mem - mem[-1];//算回原分配指针位置            delete [] mem;        }    }