c++ - 标准布局类型和 reinterpret_cast

Question

如果我已将结构的成员复制到我的类中，我是否可以从我的类转换为结构？

#include <stdint.h>
#include <sys/uio.h>

class Buffer
{
public:
    void * address;
    size_t size;

    Buffer(void * address = nullptr, size_t size = 0)
        : address(address), size(size)
    {
    }

    operator iovec *() const
    {
        // Cast this to iovec. Should work because of standard layout?
        return reinterpret_cast<iovec *>(this);
    }
}

Answer 1

首先，你不能抛弃常量:

§5.2.10p2. The reinterpret_cast operator shall not cast away constness (§5.2.11). (...)

所以你至少需要这样写

operator iovec const*() const
{
    return reinterpret_cast<iovec const*>(this);
}

或

operator iovec *()
{
    return reinterpret_cast<iovec *>(this);
}

最重要的是，您需要同时拥有 Buffer和 iovec是标准布局类型，iovec不能有比 Buffer 更严格(即更大)的对齐方式.

§5.2.10p7. An object pointer can be explicitly converted to an object pointer of a different type. When a prvalue v of type “pointer to T1” is converted to the type “pointer to cv T2”, the result is static_cast<cv T2*>(static_cast<cv void*>(v)) if both T1 and T2 are standard-layout types (§3.9) and the alignment requirements of T2 are no stricter than those of T1, or if either type is void. (...)

您还需要注意不要破坏 strict aliasing rules : 通常，您不能使用指向同一内存位置的不同类型的两个指针或引用。