Reinhard Pagitsch said:
> Hello,
>
> I have an other problem which I do not understand:
>
> Here the code piece:
>
> #define NOP 0xd3eeee
> void WriteNOP(PerlIO* fh)
> {
> WriteTriplex(fh, NOP);
> }
>
> int WriteTriplex(PerlIO* fh, long triplex)
> {
> int i;
> char buf[4] = { '\0' };
> printf("The triplex: %x\n", triplex);
> buf[0] = (char)triplex>>16;
> buf[1] = (char)triplex>>8;
> buf[2] = (char)triplex;
> printf("The triplex: %s\n", buf);
> return PerlIO_write(fh, buf, 3);
> }
>
That's a hard way to do it. You're still suffering from endianness problems
because you're assuming which byte is first, *and* you're suffering from
signed shifts.
Try this, instead:
#define NOP "\xd3\xee\xee" /* or whatever is proper C for hex char vals */
#define WRITE_CONST_STRING(perl_io, const_buffer) \
PerlIO_Write ((perl_io), (const_buffer), sizeof (const_buffer))
The benefit here is that the compiler can figure out for itself the length
parameter because the string is a compile-time constant. Since you're using
strings (uchar arrays), you have no worries about byte order. Since you're
using bytes, you have no worries about bit-shifting.
--
muppet <scott at asofyet dot org>
