in reply to Re^4: -1 from hex ffffffffffffffff ?
in thread -1 from hex ffffffffffffffff ?
Representations of numbers are always big endian.
The OP said hex string, not hex number.
To elaborate, lets take the decimal number 1000. Depending on whether it's stored big- or little-endian (in memory, in a file, etc.), you could find it as the (2-byte) string 03e8, or e803 (written in hex), and you could convert it to the number 1000 using any of the following (where 'n' and 'v' stand for "unsigned 16-bit in big-endian order" and "unsigned 16-bit in little-endian order" respectively):
print unpack('n', pack('H*', '03e8')), "\n"; # 1000 print unpack('v', reverse pack('H*', '03e8')), "\n"; # 1000 print unpack('v', pack('H*', 'e803')), "\n"; # 1000 print unpack('n', reverse pack('H*', 'e803')), "\n"; # 1000
Whereas the correct usage of 'n' or 'v' in combination with reverse does matter here, it doesn't when you have the symmetric hex string 'ffff'. Any combination will do (in particular, it's irrelevant whether you use the big- or little-endian unpack pattern):
print unpack('n', pack('H*', 'ffff')), "\n"; # 65535 print unpack('v', reverse pack('H*', 'ffff')), "\n"; # 65535 print unpack('v', pack('H*', 'ffff')), "\n"; # 65535 print unpack('n', reverse pack('H*', 'ffff')), "\n"; # 65535 print unpack('n', reverse pack('H*', 'ffff')), "\n"; # 65535 print unpack('v', pack('H*', 'ffff')), "\n"; # 65535 print unpack('v', reverse pack('H*', 'ffff')), "\n"; # 65535 print unpack('n', pack('H*', 'ffff')), "\n"; # 65535
(think of half of the 'ffff' as big-endian, and the other half as little-endian — as with 03e8/e803 above)
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Re^6: -1 from hex ffffffffffffffff ?
by ikegami (Patriarch) on Oct 02, 2009 at 19:24 UTC | |
by almut (Canon) on Oct 02, 2009 at 19:46 UTC | |
by ikegami (Patriarch) on Oct 02, 2009 at 20:15 UTC |