C近拍的定义来确定big endian还是little endian机器？

问题描述

有一个行宏定义来判断机器的字节顺序。我使用下面的code，但其转换为宏就太长了。

  unsigned char型test_endian（无效）
{
    INT test_var = 1;
    unsigned char型test_endian * =（无符号字符*）及test_var;    回报（test_endian [0] == NULL）;
}
 

解决方案

code支持任意字节命令，随时可以投入一个名为 order32.h

 的#ifndef ORDER32_H
＃定义ORDER32_H＃包括LT＆;＆limits.h中GT;
＃包括LT＆;＆stdint.h GT;＃如果CHAR_BIT！= 8
#ERROR不支持字符大小
＃万一枚举
{
    O32_LITTLE_ENDIAN = 0x03020100ul，
    O32_BIG_ENDIAN = 0x00010203ul，
    O32_PDP_ENDIAN = 0x01000302ul
};静态常量工会{unsigned char型字节[4]; uint32_t的价值; } o32_host_order =
    {{0，1，2，3}};＃定义O32_HOST_ORDER（o32_host_order.value）＃万一
 

您将通过

检查小尾数系统

  O32_HOST_ORDER == O32_LITTLE_ENDIAN
 

Is there a one line macro definition to determine the endianness of the machine. I am using the following code but converting it to macro would be too long.

unsigned char test_endian( void )
{
    int test_var = 1;
    unsigned char test_endian* = (unsigned char*)&test_var;

    return (test_endian[0] == NULL);
}

解决方案

Code supporting arbitrary byte orders, ready to be put into a file called order32.h:

#ifndef ORDER32_H
#define ORDER32_H

#include <limits.h>
#include <stdint.h>

#if CHAR_BIT != 8
#error "unsupported char size"
#endif

enum
{
    O32_LITTLE_ENDIAN = 0x03020100ul,
    O32_BIG_ENDIAN = 0x00010203ul,
    O32_PDP_ENDIAN = 0x01000302ul
};

static const union { unsigned char bytes[4]; uint32_t value; } o32_host_order =
    { { 0, 1, 2, 3 } };

#define O32_HOST_ORDER (o32_host_order.value)

#endif

You would check for little endian systems via

O32_HOST_ORDER == O32_LITTLE_ENDIAN

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