Child 的对齐小于 Base 的对齐

Question

经过深思熟虑，我将问题简化为以下简单示例：

//__declspec(align(16)) class Vec4 {}; //For testing purposes on Windows
//class Vec4 {} __attribute__((__aligned__(16))); //For testing purposes on *nix
class Base { public:
    Vec4 v; //16-byte-aligned struct
};
class Child : public Base {};
static_assert(alignof( Base)>=16,"Check 1");
static_assert(alignof(Child)>=16,"Check 2");

检查 1 次通过；检查 2 失败。 我的问题：为什么？

实际上，分配 Child（即 new Child）将导致 v 可能 8 字节对齐（v 使用 SSE ，因此这反过来会导致崩溃）。

编译器是Intel Compiler 2016，我试过g++和Clang，好像还可以。这可能是编译器错误吗？

Answer 1

根据http://en.cppreference.com/w/cpp/language/object#Alignment：

It is implementation-defined if new-expression, std::allocator::allocate, and std::get_temporary_buffer support over-aligned types. Allocators instantiated with over-aligned types are allowed to fail to instantiate at compile time, to throw std::bad_alloc at runtime, to silently ignore unsupported alignment requirement, or to handle them correctly.

作为解决方法，我建议为您的基础 class.

定义自定义 operator new

Answer 2

我有 asked Intel directly. Although it could initially be reproduced by their team, after a few days, it couldn't be. Neither I nor they had an explanation. So, magic 确实存在，没有造成任何伤害。

至于问题的实际内容：是的，看来 child 的对齐应该至少与基的对齐一样大，不这样做的编译器是错误。

N.B。堆栈分配尊重 class 的 alignof，但 malloc/new 不尊重（他们怎么知道的？）。正如另一个答案所说，必须使用自定义堆分配器。这个问题与正确的 alignof 本身是分开的。

Child 的对齐小于 Base 的对齐

Alignment of Child is Smaller Than Alignment of Base

c++

inheritance

alignment

icc