gcc中的位域字节序 [英] Bitfield endianness in gcc

问题描述

位域的字节序由实现定义。有没有一种方法可以在编译时通过宏或其他编译器标志检查gcc的位域字节序实际上是什么？

The endianness of bitfields is implementation defined. Is there a way to check, at compile time, whether via some macro or other compiler flag, what gcc's bitfield endianness actually is?

换句话说，给定类似以下内容：

In other words, given something like:

struct X {
    uint32_t a : 8;
    uint32_t b : 24;
};

我是否有办法在编译时知道 a 是 X 中的第一个字节还是最后一个字节？

Is there a way for me to know at compile time whether or not a is the first or last byte in X?

推荐答案

在Linux系统上，您可以检查 __ BYTE_ORDER 宏以查看如果是 __ LITTLE_ENDIAN 或 __ BIG_ENDIAN 。

On Linux systems, you can check the __BYTE_ORDER macro to see if it is __LITTLE_ENDIAN or __BIG_ENDIAN. While this is not authoritative, in practice it should work.

在 struct的定义中，暗示这是正确的方法netinet / ip.h中的iphdr ，用于IP标头。第一个字节包含两个4位字段，这些字段被实现为位字段，因此顺序很重要：

A hint that this is the right way to do it is in the definition of struct iphdr in netinet/ip.h, which is for an IP header. The first byte contains two 4-bit fields which are implemented as bitfields, so the order is important:

struct iphdr
  {
#if __BYTE_ORDER == __LITTLE_ENDIAN
    unsigned int ihl:4;
    unsigned int version:4;
#elif __BYTE_ORDER == __BIG_ENDIAN
    unsigned int version:4;
    unsigned int ihl:4;
#else
# error "Please fix <bits/endian.h>"
#endif
    u_int8_t tos;
    u_int16_t tot_len;
    u_int16_t id;
    u_int16_t frag_off;
    u_int8_t ttl;
    u_int8_t protocol;
    u_int16_t check;
    u_int32_t saddr;
    u_int32_t daddr;
    /*The options start here. */
  };

这篇关于gcc中的位域字节序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！