gcc是否提供内置读取大端内存？(Does gcc provide builtin to read big endian memory?)

gcc是否提供内置读取大端内存？(Does gcc provide builtin to read big endian memory?)

我需要从两个小端核（例如arm）读/写设备寄存器（在我的情况下是在大端地址空间中）。 我不想使用中间端转换例程（例如htonl）。 我的理解是现代核心具有指令，使用它们可以直接读取/写入大端存储器。

gcc是否为上述内容提供了自动生成正确汇编指令的内置函数？

I need to read/write device registers (which in my case are in big endian address space) from both little endian cores (e.g. arm). I do not want to use intermediate endian conversion routines (e.g. htonl). My understanding is that modern cores have instructions using which they can do reads/writes to big endian memory directly.

Does gcc offer any builtin for the above which automatically generates right assembly instruction(s)?

最满意答案

GCC 4.3.0提供内置功能：

内置函数：int32_t __builtin_bswap32（int32_t x），它返回逆转的顺序。 例如。 11223344将是0x44332211 内置函数：int64_t __builtin_bswap64（int64_t x）也类似于__builtin_bswap32，但它返回64位。

请参阅内置的小心 。

There are built-in functions available from GCC 4.3.0:

Built-in Function: int32_t __builtin_bswap32 (int32_t x), it returns the order of byes reversed. for eg. 11223344 will be 0x44332211 Built-in Function: int64_t __builtin_bswap64 (int64_t x) is also similar to __builtin_bswap32, except that it returns 64 bits.

Please refer to the Beware the builtins.