On 2019-11-01 12:56:57 -0400, Jeff King wrote:
> On Fri, Nov 01, 2019 at 03:25:10PM +0700, Doan Tran Cong Danh wrote:
> > + v = (unsigned char)(out[0]) + (unsigned char)(out[1]);
> > + return v != 0xfe + 0xff;
>
> I wondered at first why not just:
>
> return v[0] != 0xfe || v[1] != 0xff;
>
> but the answer is that you are catching BOMs of either endianness. We
> might at some point need to distinguish the two,
Yes, it's about catching BOM of either endianess. To distinguish
between LE and BE, it'll be better to use AC_C_BIGENDIAN instead (it's
friendlier for cross-compiling).
Anyway, we couldn't compare out[0] and out[1] with 0xf{e,f} directly
because char could be either signed (-2 and -1) or unsigned (0xfe, 0xff).
We must cast it to unsigned char (required to be 8 bits by POSIX),
then let the integer promotion promote it to int (at least 16 bits).
--
Danh
