c++ - std::aligned_storage 的 static_cast 和 reinterpret_cast

Question

有人可以解释一下有关在http://en.cppreference.com/w/cpp/types/aligned_storage中进行转换的代码吗？

可以下面的代码

return *static_cast<const T*>(static_cast<const void*>(&data[pos]));

被替换为

 return *reinterpret_cast<const T*>(&data[pos]);

?

为什么这里使用两个铸件？非常感谢。

洪

Answer 1

According to the standard (§ 5.2.10 reinterpret_cast, section 7):

A pointer to an object can be explicitly converted to a pointer to a different object 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 and the alignment requirements of T2 are no stricter than those of T1.

Converting a prvalue of type “pointer to T1” to the type “pointer to T2” (where T1 and T2 are object types and where the alignment requirements of T2 are no stricter than those of T1) and back to its original type yields the original pointer value. The result of any other such pointer conversion is unspecified.

So, we could make the following conclusion:

1. reinterpret_cast<*T>(ptr) is eqiuvalent to static_cast<*T>(static_cast<void*>(ptr))
2. static_cast<>(ptr) is not always equal to ptr, but reinterpret_cast<>(ptr) is always equal to ptr
3. if there is no alignment issues, we can use reinterpret_cast safely