内容简介：http://stackoverflow.com/questions/27003727/does-this-really-break-strict-aliasing-rules
当我用g编译这个示例代码时,我得到这个警告：
warning: dereferencing type-punned pointer will break strict-aliasing rules [-Wstrict-aliasing]
代码：
#include <iostream>
int main()
{
alignas(int) char data[sizeof(int)];
int *myInt = new (data) int;
*myInt = 34;
std::cout << *reinterpret_cast<int*>(data);
}
在这种情况下,数据别名不是int,因此将其返回到int不会违反严格的别名规则？还是我在这里遗漏的东西？
编辑：奇怪,当我定义这样的数据：
alignas(int) char* data = new char[sizeof(int)];
编译器警告消失了.堆栈分配是否与严格的混叠有所不同？事实上,它是一个char []而不是一个char *意味着它不能实际上任何类型的别名？
：
If, after the lifetime of an object has ended and before the storage
which the object occupied is reused or released, a new object is
a
, a reference that referredthe name of the original object will
and, once the lifetime of thenew object has started, can be used to manipulate the new object, if :
(7.1) — [..]
the new object is of the same type as the
original object (ignoring the top-level cv-qualifiers)
,新对象不是一个char数组,而是一个int. P0137 ,正式化了指点的概念,添加了洗衣：
[ Note : If these conditions are not met, a pointer to the new object
can be obtained from a pointer that represents the address of its
storage by calling std::launder
(18.6 [support.dynamic]). — end note
]
即您的代码段可以这样纠正：
std::cout << *std::launder(reinterpret_cast<int*>(data));
..或者只是从放置新的结果初始化一个新的指针,这也消除了警告.
http://stackoverflow.com/questions/27003727/does-this-really-break-strict-aliasing-rules
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