Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
I need to convert a number like this: -3205.0569059
... into an 8-byte double (big and little endian), e.g. 4f 3e 52 00 2a bc 93
d3 (I just made up those 8 byte values).
Is this easy in perl? Are long and short ints easy as well?
$ perl -le'print unpack "H*", pack "d", -3205.0569059'
e626c5221d0aa9c0
Maybe this is just my own ignorance on big-endian vs. little endian, but
this code:
print "big-endian: ", unpack("H*", pack("d", -3205.0569059)), "\n";
print "little-endian: ", unpack("h*", pack("d", -3205.0569059)), "\n";
prints:
big-endian: e626c5221d0aa9c0
little-endian: 6e625c22d1a09a0c
... when I expected the little endian to look more like:
c0 a9 0a 1d 22 c5 26 e6 (spacing for readability)
Did I do it wrong (i.e. is "h*" the wrong string?), or am I confused on how
big vs. little endian works?
The difference between 'h' and 'H' has nothing to do with endianness:
perldoc -f pack
[ SNIP ]
h A hex string (low nybble first).
H A hex string (high nybble first).
A nybble is half of a byte so the only thing exchanged is the order of
each byte's two halves.
AFAIK floating point numbers don't have endianness (but I may be wrong.)
(Besides, in the above example, endianness would be on the pack('d')
side of the equation.)
John
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