确定C ++中的最大可能对齐 [英] Determining maximum possible alignment in C++
问题描述
有没有任何便携式方式来确定任何类型的最大可能对齐方式是什么?
Is there any portable way to determine what the maximum possible alignment for any type is?
在x86上的示例,SSE指令需要16字节对齐,但据我所知,没有指令需要更多,所以任何类型都可以安全地存储到16字节对齐缓冲区。
For example on x86, SSE instructions require 16-byte alignment, but as far as I'm aware, no instructions require more than that, so any type can be safely stored into a 16-byte aligned buffer.
我需要创建一个缓冲区(例如一个char数组),我可以写任意类型的对象,所以我需要能够依赖于开始
I need to create a buffer (such as a char array) where I can write objects of arbitrary types, and so I need to be able to rely on the beginning of the buffer to be aligned.
如果所有其他都失败,我知道分配一个具有 new
的字符数组是有保证的具有最大对齐,但是使用TR1 / C ++ 0x模板 alignment_of
和 aligned_storage
,我想知道是否可以在我的缓冲区类中创建缓冲区,而不需要动态分配的数组的额外指针间接。
If all else fails, I know that allocating a char array with new
is guaranteed to have maximum alignment, but with the TR1/C++0x templates alignment_of
and aligned_storage
, I am wondering if it would be possible to create the buffer in-place in my buffer class, rather than requiring the extra pointer indirection of a dynamically allocated array.
想法?
我知道有很多选项用于确定有限类型集合的最大对齐:A union,或者只是来自TR1的 alignment_of
,但我的问题是类型集是无限的。我不知道哪些对象必须存储到缓冲区中。
I realize there are plenty of options for determining the max alignment for a bounded set of types: A union, or just alignment_of
from TR1, but my problem is that the set of types is unbounded. I don't know in advance which objects must be stored into the buffer.
推荐答案
在C ++ 0x中, Align
模板参数 std :: aligned_storage< Len,Align>
的默认参数为default-alignment定义为(N3225§20.7.6.6表56):
In C++0x, the Align
template parameter of std::aligned_storage<Len, Align>
has a default argument of "default-alignment," which is defined as (N3225 §20.7.6.6 Table 56):
默认对齐的值应是任何C ++对象类型大小不大于
Len
。
清除是否将SSE类型视为C ++对象类型。
It isn't clear whether SSE types would be considered "C++ object types."
默认参数不是TR1的一部分 aligned_storage
;它被添加为C ++ 0x。
The default argument wasn't part of the TR1 aligned_storage
; it was added for C++0x.
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