预处理器为m68k定义和宏？

Question

我们Debian's testing/build infrastructure下m68k的失败构建：

config.h:29:3: error: #error "IS_LITTLE_ENDIAN is set, but __BYTE_ORDER__ does not equal __ORDER_LITTLE_ENDIAN__" 
# error "IS_LITTLE_ENDIAN is set, but __BYTE_ORDER__ does not equal __ORDER_LITTLE_ENDIAN__" 

解决方法是容易的，但我需要知道预处理器定义了平台。我没有架构的机器，所以我不能将它们转储到gcc -dM -E - </dev/null | sort。

有人可以提供m68k的预处理器输出吗？

Answer 1

如果你愿意消磨时间，你可以为该平台构建一个交叉编译器。如果您只有关心那一个特定的标志，请尝试将此测试添加到您的configure.ac：

#include <stdint.h> 

enum { 
    ENDIAN_UNKNOWN, 
    ENDIAN_BIG, 
    ENDIAN_LITTLE, 
    ENDIAN_BIG_WORD, /* Middle-endian, Honeywell 316 style */ 
    ENDIAN_LITTLE_WORD /* Middle-endian, PDP-11 style */ 
}; 

int endianness(void) 
{ 
    union 
    { 
    uint32_t value; 
    uint8_t data[sizeof(uint32_t)]; 
    } number; 

    number.data[0] = 0x00; 
    number.data[1] = 0x01; 
    number.data[2] = 0x02; 
    number.data[3] = 0x03; 

    switch (number.value) 
    { 
    case UINT32_C(0x00010203): return ENDIAN_BIG; 
    case UINT32_C(0x03020100): return ENDIAN_LITTLE; 
    case UINT32_C(0x02030001): return ENDIAN_BIG_WORD; 
    case UINT32_C(0x01000302): return ENDIAN_LITTLE_WORD; 
    default:     return ENDIAN_UNKNOWN; 
    } 
}