On Fri, 25 Jul 2003 14:57:57 +0200, Alexander Eisenhuth <st****@stacom-software.de> wrote:
Hallo all there,
maby I don't see the forest because of all that trees, but :
from struct import *
# how can I convert
f = float(0.5)
# with
bin_str = pack('!f', 0.5)
# now bin_str is '?\x00\x00\x00'
# to "3F000000" ?????
That leading '?' is the character representation of a string byte whose ordinal value
is 3F hex, followed by three characters represented by \x00 (whose ordinal
values are zero). I.e., your binary bytes are being represented as characters in a string,
and when it's shown interactively you see the values as a sequence of character glyphs
(or \x.. hex escape representations if no glyphs are available), which is the normal way
to display a string.
If you want the *hex string* representation of that string-represented byte sequence, you can
convert each byte-as-character first to its ordinal value and then the ordinal value to a
2-char hex string representation, and then join those all into a single string,
e.g.,, showing the character transformations:
[c for c in struct.pack('!f', 0.5)]
['?', '\x00', '\x00', '\x00'] [ord(c) for c in struct.pack('!f', 0.5)]
[63, 0, 0, 0] ['%02X'%ord(c) for c in struct.pack('!f', 0.5)]
['3F', '00', '00', '00']
And finally, all you need is
''.join(['%02X'%ord(c) for c in struct.pack('!f', 0.5)])
'3F000000'
Now you have a hex string representation of the network-ordered (big-endian-ordered)
bytes of a single precision (32-bit) float.
Regards,
Bengt Richter
