On Wed, Sep 23, 2009 at 15:32, Bob McConnell <r...@cbord.com> wrote:
> From: Uri Guttman
>
>>>>>>> "BM" == Bob McConnell <r...@cbord.com> writes:
>>
>> BM> From: Bryan R Harris
>> >>
>> >> 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
>> BM> bc 93
>> >> d3 (I just made up those 8 byte values).
>> >>
>> >> Is this easy in perl? Are long and short ints easy as well?
>>
>> BM> The sprintf() family is your friend.
>>
>> that will only generate text (hex and other formats). he needs pack
>> which does exactly what he wants. read perlpacktut for a tutorial on
>> pack/unpack and then perlfunc -f pack for the reference on it.
>
> That statement just confuses me. His initial value of -3205.0569059 is
> also text. It is the human readable representation of the number, and is
> not anything like what it looks like inside the computer. He just asked
> for a different format for that text. Why is sprintf not a reasonable
> way to do that?
snip
Because sprintf "%16x", .5 returns "0000000000000000" and you can't
recover .5 from that. On the other hand you can say:
#!/usr/bin/perl
use strict;
use warnings;
my $n = -3205.0569059;
my $packed = pack "d", $n;
my $hex = unpack "H*", $packed;
my $unpacked = unpack "d", pack "H*", $hex;
print
"n = $n\n",
"packed = $packed\n",
"hex = $hex\n",
"unpacked = $unpacked\n";
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
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