如何对齐指针

如何将指针与 16 字节边界对齐？

我找到了这段代码，不确定是否正确

char* p= malloc(1024);

if ((((unsigned long) p) % 16) != 0) 
{
     unsigned char *chpoint = (unsigned char *)p;        
     chpoint += 16 - (((unsigned long) p) % 16);
     p = (char *)chpoint;
}

这行得通吗？

thanks

C++0x提出std::align http://en.cppreference.com/w/cpp/memory/align，它就是这样做的。

// get some memory
T* const p = ...;
std::size_t const size = ...;

void* start = p;
std::size_t space = size;
void* aligned = std::align(16, 1024, p, space);
if(aligned == nullptr) {
    // failed to align
} else {
    // here, p is aligned to 16 and points to at least 1024 bytes of memory
    // also p == aligned
    // size - space is the amount of bytes used for alignment
}

这看起来非常低级。我认为

// also available in Boost flavour
using storage = std::aligned_storage_t<1024, 16>;
auto p = new storage;

也有效。如果您不小心，您很容易违反别名规则。如果您心里有一个精确的场景（在 16 字节边界上容纳 N 个 T 类型的对象？）我想我可以推荐一些更好的东西。