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

[download]

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

[download]

(think of half of the 'ffff' as big-endian, and the other half as little-endian — as with 03e8/e803 above)

In reply to Re^5: -1 from hex ffffffffffffffff ? by almut
in thread -1 from hex ffffffffffffffff ? by Spooky

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