c++ - 使用placement new操作符时我真的需要担心对齐吗？

Question

我读到了 When should I worry about alignment?但我仍然不知道我是否必须担心放置 new 运算符返回的未对齐指针 - 就像在这个例子中一样:

class A {
public:
   long double a;
   long long b;
   A() : a(1.3), b(1234) {}
};

char buffer[64];

int main() {
   // (buffer + 1) used intentionally to have wrong alignment
   A* a = new (buffer + 1) A(); 
   a->~A();
}

__alignof(A) == 4，(buffer + 1) 未与 4 对齐。但一切正常——这里有完整的例子:http://ideone.com/jBrk8

如果这取决于架构，那么我使用的是:linux/powerpc/g++ 4.x.x。

[更新] 在发布这个问题后，我阅读了这篇文章:http://virtrev.blogspot.de/2010/09/memory-alignment-theory-and-c-examples.html .也许在我的情况下唯一的缺点是性能损失，我的意思是未对齐的访问成本高于对齐？

Answer 1

当你在缓冲区上调用placement new时:

A *a = new (buf) A;

您正在调用内置的 void* operator new (std::size_t size, void* ptr) noexcept定义在:

c++11

18.6.1.3 Placement forms [new.delete.placement]

These functions are reserved, a C++ program may not define functions that displace the versions in the Standard C++ library (17.6.4). The provisions of (3.7.4) do not apply to these reserved placement forms of operator new and operator delete.

void* operator new(std::size_t size, void* ptr) noexcept;
Returns: ptr.
Remarks: Intentionally performs no other action.

(3.7.4) 的规定 包括返回的指针应该适当对齐，所以它适用于 void* operator new (std::size_t size, void* ptr) noexcept如果传入一个指针，则返回一个未对齐的指针。不过，这不会让 你 摆脱困境:

5.3.4 New [expr.new]

[14] Note: when the allocation function returns a value other than null, it must be a pointer to a block of storage in which space for the object has been reserved. The block of storage is assumed to be appropriately aligned and of the requested size.

因此，如果您将未对齐的存储传递给放置新表达式，则您违反了存储已对齐的假设，结果是 UB。

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确实，在您上面的程序中，如果您替换 long long b与 __m128 b (在 #include <xmmintrin.h> 之后)然后程序会出现段错误，正如预期的那样。