为什么必须在 typedef 中指定此对齐属性？

Question

我最初是在我的平板电脑上写这个问题的，并且在这样做时采取了很多捷径，所以我认为最终会导致阅读 and/or 试图回答这个问题的人感到困惑。

我不是在寻求解决我最初遇到的问题的方法。如果您真的想了解背景故事，请阅读下一段，否则请跳过。

导致此问题的原因是一些旧代码对 {struct, data, struct, data, ...} 形式的数据数组进行操作，其中每个 data 具有任意长度。代码通过指针访问每个结构，当我们切换到 gcc 时，由于访问未对齐，它开始在 Solaris 中崩溃。解决这个问题的一个想法是改变类型的对齐方式，如下所示，但我可能不会那样做。

要回答的问题可以概括为：

文档指出无法使用 aligned 减少对齐，但我可以使用 typedef 来做到这一点。它是否按预期工作？
如果它按预期工作，为什么它需要 typedef？为什么我不能降低对齐作为结构定义的一部分？
- 注意: typedef struct {...}__attribute__((aligned(1))) Typename;也可以

这是 link to some sample code running on wandbox。如果 link 死机：

#include <cstdio>
#include <assert.h>

#define ALIGN __attribute__((aligned(1)))

struct       Misaligned_1_t { int x; double y; float z; };
struct ALIGN Misaligned_2_t { int x; double y; float z; };
struct       Misaligned_3_t { int x; double y; float z; } ALIGN;

// The gcc documentation indicates that the "aligned" attribute
// can only be used to increase alignment, so I was surprised
// to discover this actually works.  Why does it work?
typedef Misaligned_1_t ALIGN Aligned_t;

int main( int, char** ) {
  char buffer[256];
  // The following is meant to simulate a more complicated scenario:
  //   {SomeStruct, char[arbitrary length], SomeStruct, char[arbitrary length], ...}
  // ... where accessing, using and changing each SomeStruct will result in
  // misaligned accesses.
  auto *m1 = (Misaligned_1_t*)&buffer[1];
  auto *m2 = (Misaligned_1_t*)&buffer[1];
  auto *m3 = (Misaligned_1_t*)&buffer[1];
  auto *a1 = (Aligned_t*)&buffer[1];

  // The documentation says we can only reduce alignment with the "packed" attribute,
  // but that would change the size/layout of the structs.  This is to demonstrate
  // that each type is the same size (and should have the same layout).
  assert(   sizeof(m1) == sizeof(m2)
         && sizeof(m1) == sizeof(m3)
         && sizeof(m1) == sizeof(a1) );

  m1->y = 3.14159265358979323846264; // misaligned access

  std::printf( "%0.16f\n", m2->y ); // misaligned access
  std::printf( "%0.16f\n", m3->y ); // misaligned access
  std::printf( "%0.16f\n", a1->y ); // works fine

  return 0;
}

Answer 1

来自 gcc help files

You may specify the aligned and transparent_union attributes either in a typedef declaration or just past the closing curly brace of a complete enum, struct or union type definition and the packed attribute only past the closing brace of a definition.

所以你可以使用

struct Test_t {
  int x;
  double y;
  float z;
} __attribute__((aligned(1)));

然后用

定义变量

struct Test_t a,b;
struct Test_t *test;

或者你可以使用上面给出的方式。是一样的。

Answer 2

我找到了答案。我一定是瞎了。来自 GCC 文档：

When used on a struct, or struct member, the aligned attribute can only increase the alignment; in order to decrease it, the packed attribute must be specified as well. When used as part of a typedef, the aligned attribute can both increase and decrease alignment, and specifying the packed attribute generates a warning.

为什么必须在 typedef 中指定此对齐属性？

Why does this alignment attribute have to be specified in a typedef?

c

attributes

memory-alignment

struct-member-alignment