C++结构对齐问题
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1455458/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
C++ struct alignment question
提问by NJChim
I have a predefined struct (actually several) where variables span across 32-bit word boundary. In Linux (and Windows using GCC) I am able to get my structs to pack to the correct size using 'attribute((packed))'. However I cannot get it to work the same way using VC++ and #pragma pack.
我有一个预定义的结构（实际上是几个），其中变量跨越 32 位字边界。在 Linux（和使用 GCC 的 Windows）中，我能够使用“attribute((packed))”将我的结构打包到正确的大小。但是，我无法使用 VC++ 和 #pragma pack 使其以相同的方式工作。
Using GCC this returns a correct size of 6 bytes:
使用 GCC 这将返回 6 个字节的正确大小：
struct
{
unsigned int a : 3;
unsigned int b : 1;
unsigned int c : 15;
unsigned int troubleMaker : 16;
unsigned short padding : 13;
} __attribute__((packed)) s;
Using VC++ this returns an incorrect size of 8 bytes
使用 VC++ 这会返回错误的 8 字节大小
#pragma pack(push)
#pragma pack(1)
struct
{
unsigned int a : 3;
unsigned int b : 1;
unsigned int c : 15;
unsigned int troubleMaker : 16;
unsigned short padding : 13;
} s;
#pragma pack(pop)
I can get things to work by splitting 'troubleMaker' across the boundary manually but I'd prefer not to. Any ideas?
我可以通过手动跨边界拆分“troubleMaker”来使事情发挥作用，但我不想这样做。有任何想法吗？
回答by DigitalRoss
Crazy idea: just write a C99 or C++03 -conforming program in the first place
疯狂的想法：首先只写一个符合 C99 或 C++03 的程序
I would suggest not using vendor-specific C language extensions to match device or network bit formats. Even if you get the fields to line up using a series of one-per-vendor language extensions, you still have byte order to worry about, and you still have a struct layout that requires extra instructions to access.
我建议不要使用特定于供应商的 C 语言扩展来匹配设备或网络位格式。即使您使用一系列每个供应商的语言扩展来排列字段，您仍然需要担心字节顺序，并且您仍然有一个需要额外指令才能访问的结构布局。
You can write a C99 conforming program that will work on any architecture or host and at maximum speed and cache efficiency by using the standardized C API string and memory copy functions and the Posix hton and ntoh functions.
通过使用标准化的 C API 字符串和内存复制函数以及 Posix hton 和 ntoh 函数，您可以编写符合 C99 的程序，该程序可以在任何架构或主机上运行，并以最大速度和缓存效率运行。
A good practice is to use the following functions for which there exist published standards:
一个好的做法是使用以下已发布标准的函数：
C99: memcpy(), Posix: htonl(), htons(), ntohl(), ntohs()
Update:here is some code that should work the same everywhere. You may need to get <stdint.h>from this projectif Microsoft stillhasn't implemented it for C99, or just make the usual assumptions about int sizes.
更新：这里有一些代码应该在任何地方都一样。如果微软还没有为 C99 实现它，你可能需要<stdint.h>从这个项目中得到它，或者只是对 int 大小做出通常的假设。
#include <stdlib.h>
#include <stdint.h>
#include <string.h>
#include <stdio.h>
#include <arpa/inet.h>
struct packed_with_bit_fields { // ONLY FOR COMPARISON
unsigned int a : 3;
unsigned int b : 1;
unsigned int c : 15;
unsigned int troubleMaker : 16;
unsigned short padding : 13;
} __attribute__((packed)); // USED ONLY TO COMPARE IMPLEMENTATIONS
struct unpacked { // THIS IS THE EXAMPLE STRUCT
uint32_t a;
uint32_t b;
uint32_t c;
uint32_t troubleMaker;
}; // NOTE NOT PACKED
struct unpacked su;
struct packed_with_bit_fields sp;
char *bits = "Lorem ipsum dolor";
int main(int ac, char **av) {
uint32_t x; // byte order issues ignored in both cases
// This should work with any environment and compiler
memcpy(&x, bits, 4);
su.a = x & 7;
su.b = x >> 3 & 1;
su.c = x >> 4 & 0x7fff;
memcpy(&x, bits + 2, 4);
su.troubleMaker = x >> 3 & 0xffff;
// This section works only with gcc
memcpy(&sp, bits, 6);
printf( sp.a == su.a
&& sp.b == su.b
&& sp.c == su.c
&& sp.troubleMaker == su.troubleMaker
? "conforming and gcc implementations match\n" : "huh?\n");
return 0;
}
回答by Paul Lalonde
Alignment and ordering of bitfields are notoriously implementation-specific. It is muchsafer to declare a normal integer field and manipulate the "bitfields" within using masks and bitwise (| & ^) operators .
众所周知，位域的对齐和排序是特定于实现的。这是很多安全声明一个正常的整型字段和操作使用内口罩和按位“位域”（|＆^）运营商。
回答by JaredPar
I don't believe this behavior is supported in Visual Studio. In addiction to the pack macro I tried using __declspec(align(1))and got the same behavior. I think you are stuck with 12 bytes or re-ordering your structure a bit.
我认为 Visual Studio 不支持这种行为。沉迷于包宏，我尝试使用__declspec(align(1))并得到了相同的行为。我认为你被 12 个字节困住了，或者对你的结构重新排序了一下。
回答by DigitalRoss
Shorter example with only conforming code
只有符合代码的较短示例
struct unpacked { // apparently my other example was too long and confusing
uint32_t a; // ...here is a much shorter example with only the conforming
uint32_t b; // ...code. (The other program had the gcc-specific declaration,
uint32_t c; // but only for test code. Still, it was a bit long.)
uint32_t troubleMaker;
};
struct unpacked su;
char *bits = "Lorem ipsum dolor";
void f(void) {
uint32_t x;
memcpy(&x, bits, 4);
su.a = x & 7;
su.b = x >> 3 & 1;
su.c = x >> 4 & 0x7fff;
memcpy(&x, bits + 2, 4);
su.troubleMaker = x >> 3 & 0xffff;
return 0;
}
回答by Anton Tykhyy
I believe VC++ doesn't support this, and I have grave doubts whether GCC's behaviour in this respect is actually standard.
我相信 VC++ 不支持这一点，我严重怀疑 GCC 在这方面的行为是否真的是标准的。
回答by Goz
If it absoloutely defnitely needs to be 6 bytes then define it as 3 shorts and get the data out yourself ... it won't slow things down ... the compiler is just doing this anyway ...
如果它绝对需要是 6 个字节，那么将它定义为 3 个 shorts 并自己获取数据......它不会减慢速度......无论如何编译器只是这样做......
