Messages in this thread | | | Date | Wed, 30 May 2007 11:24:36 +0530 | From | "Nitin Gupta" <> | Subject | Re: [RFC] LZO de/compression support - take 6 |
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On 5/30/07, Jan Engelhardt <jengelh@linux01.gwdg.de> wrote: > > On May 28 2007 20:04, Nitin Gupta wrote: > > > > * Changelog vs. original LZO: > > 1) Used standard/kernel defined data types: (this eliminated _huge_ > > #ifdef chunks) > > lzo_bytep -> unsigned char * > > lzo_uint -> size_t > > lzo_xint -> size_t > > Is this safe (as far as compressed LZO stream is concerned) -- > or is it even needed (could it be unsigned int)? > > > - m_pos -= (*(const unsigned short *)ip) >> 2; > > -#else > > - m_pos = op - 1; > > - m_pos -= (ip[0] >> 2) + (ip[1] << 6); > > -#endif > > > > + m_pos = op - 1 - (cpu_to_le16(*(const u16 *)ip) >> 2); > > > > (Andrey suggested le16_to_cpu for above but I think it should be cpu_to_le16). > > *** Need testing on big endian machine *** > > On i386, both cpu_to_le16 and le16_to_cpu do nothing. > On sparc for example, cpu_to_leXX and leXX_to_cpu do 'the same' ;-) > they swap 1234<->4321. > > It is the bytestream (ip) that is reinterpreted as uint16_t. > And I really doubt that the LZO author has a big-endian machine, > given the days of ubiquitous x86.
> le16_to_cpu it is. >
I just missed the point that leXX_to_cpu() and cpu_to_leXX() are identical on BE machine anyway. But then why you think it should be le_16_cpu() -- how will this make any difference?
For your ref (from big_endian.h): #define __cpu_to_le16(x) ((__force __le16)__swab16((x))) #define __le16_to_cpu(x) __swab16((__force __u16)(__le16)(x))
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