The last time I checked, nothing needs to be done to a
scalar variable containing a string of digit characters
in order for it to be treated as an integer. The fact that
it is made up of digits means that it will be interpreted
as a numeric value in any sort of numeric context -- while
still being used as a string in any string context.
Run this snippet and see what you get:
$data = '01142000';
print "$data + 1 = ", $data + 1, $/;
print "$data / 2 = ", $data / 2, $/;
print "sqrt($data) = ", sqrt( $data ), $/;
The cases where you really need unpack (and pack) tend to
be hard to explain... and builtins like "ord()", "chr()",
"hex()" and so on tend to provide the most common functions
that (un)pack might be used for -- and in much clearer
terms.
BTW, I think the integer formats you were trying (I7,i7,
L7,l7) are supposed to tell unpack that the scalar data
being passed to it should contain seven binary "int" (i/I)
or "long int" (l/L) values -- these would be either 16- and
32-bit values or 32- and 64-bit values (or 32- and 32-bit
values), depending on your
system (and maybe depending on how your perl interpreter was
compiled for your system). So, it would have tried to read
7*2=14 bytes (16-bit i/I) or 7*4=28 bytes (32-bit i/I or l/L)
or 7*8=56 bytes (64-bit l/L) from $data. But you only fed it a scalar
containing 8 bytes. Whatever it ended up doing, at least
it did the same thing consistently for these cases.
To top it off, of course, the perldoc descriptions for
of pack and unpack are undeniable the hardest to grasp.
Don't feel bad if you don't grok it, even after working
through your own experiments with it.
(update: fixed a couple typos) | [reply]
[d/l] |
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[d/l] |
chr '0' '1' '1' '4' '2' '0' '0'
ord 48 49 49 52 50 48 48
intel word lo lh hl hi lo lh hl ??
Now the A7 template unpacks this string as its ascii characters, just as it was constructed. On the ia-32 architecture native int and native long are both customarily 4 bytes, so you expect i,l to give the same results, as well as I,L, the unsigned variants. If the number is less than 2**31, unsigned, both signed and unsigned give the same results.
Now unpack's repeat count templates give up if the data string is too short to provide all the fields. '0114200' only has enough data for one native int, so that's all you get:
perl -e'print 52*2**24 + 49*2**16 + 49*2**8 + 48,$/;"
875639088
After Compline,
Zaxo | [reply]
[d/l] |
int is probably what you are after...
I recommend that you open up the perlfunc man page (man perlfunc or perldoc perlfunc on your system) and get familiar with all of the built-ins.
Update:graff is of course right a scalar consisting of only digits is already an int or can be... /me wanders off to re-read perldata :)
--
Until you've lost your reputation, you never realize what a burden it was or what freedom really is. -Margaret Mitchell
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