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Sandro CAZZANIGA
Nov 16, 2011, 8:26:39 AM11/16/11
to f...@perl.org
hi!
Just a little JAPH for convert decimal to binary.... Feel free to comment
it ;)
----------------------------------------------------------------
#!/usr/bin/perl
use strict;
use warnings;
eval {@ARGV} or die("No args") ;
foreach my $arg(@ARGV) {
my $convert = unpack("B*",pack("N", $arg));
my @convertsp = split("", $convert);
my $count = 0;
foreach (@convertsp) {
unless ($convertsp[$count] == 1) {
$convertsp[$count] = "" ;
$count++;
}
}
print("$arg: @convertsp \n");
}
exit 0;
Just a little JAPH for convert decimal to binary.... Feel free to comment
it ;)
----------------------------------------------------------------
#!/usr/bin/perl
use strict;
use warnings;
eval {@ARGV} or die("No args") ;
foreach my $arg(@ARGV) {
my $convert = unpack("B*",pack("N", $arg));
my @convertsp = split("", $convert);
my $count = 0;
foreach (@convertsp) {
unless ($convertsp[$count] == 1) {
$convertsp[$count] = "" ;
$count++;
}
}
print("$arg: @convertsp \n");
}
exit 0;
Georg Moritz
Nov 16, 2011, 9:32:03 AM11/16/11
to Sandro CAZZANIGA, f...@perl.org
Greetings Sandro & all,
From the keyboard of Sandro CAZZANIGA [16.11.11,14:26]:
> hi!
>
> Just a little JAPH for convert decimal to binary.... Feel free to comment
> it ;)
golfed down a bit, just for fun...
#!/usr/bin/perl -l
@ARGV or die "No args";
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0+//?$_:$_ for@ARGV;
> ----------------------------------------------------------------
> #!/usr/bin/perl
> use strict;
> use warnings;
> eval {@ARGV} or die("No args") ;
> foreach my $arg(@ARGV) {
> my $convert = unpack("B*",pack("N", $arg));
> my @convertsp = split("", $convert);
> my $count = 0;
> foreach (@convertsp) {
> unless ($convertsp[$count] == 1) {
> $convertsp[$count] = "" ;
> $count++;
> }
> }
> print("$arg: @convertsp \n");
> }
> exit 0;
>
cheers,
0--gg-
--
_($_=" "x(1<<5)."?\n".q·/)Oo. G°\ /
/\_¯/(q /
---------------------------- \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
From the keyboard of Sandro CAZZANIGA [16.11.11,14:26]:
> hi!
>
> Just a little JAPH for convert decimal to binary.... Feel free to comment
> it ;)
#!/usr/bin/perl -l
@ARGV or die "No args";
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0+//?$_:$_ for@ARGV;
> ----------------------------------------------------------------
> #!/usr/bin/perl
> use strict;
> use warnings;
> eval {@ARGV} or die("No args") ;
> foreach my $arg(@ARGV) {
> my $convert = unpack("B*",pack("N", $arg));
> my @convertsp = split("", $convert);
> my $count = 0;
> foreach (@convertsp) {
> unless ($convertsp[$count] == 1) {
> $convertsp[$count] = "" ;
> $count++;
> }
> }
> print("$arg: @convertsp \n");
> }
> exit 0;
>
0--gg-
--
_($_=" "x(1<<5)."?\n".q·/)Oo. G°\ /
/\_¯/(q /
---------------------------- \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}
John Douglas Porter
Nov 16, 2011, 9:32:35 AM11/16/11
to f...@perl.org
What's the point? Seems to me the only purpose of this code beyond the obvious
unpack("B*",pack("N", $arg))
line is to reformat the result into an odd, probably application-dependent format.
--- On Wed, 11/16/11, Sandro CAZZANIGA <cazzanig...@gmail.com> wrote:
Ronald J Kimball
Nov 16, 2011, 9:57:34 AM11/16/11
to Georg Moritz, Sandro CAZZANIGA, f...@perl.org
On Wed, Nov 16, 2011 at 03:32:03PM +0100, Georg Moritz wrote:
> golfed down a bit, just for fun...
>
> #!/usr/bin/perl -l
> @ARGV or die "No args";
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0+//?$_:$_ for@ARGV;
That should be s/0*//, otherwise you'll get the wrong result for numbers
> golfed down a bit, just for fun...
>
> #!/usr/bin/perl -l
> @ARGV or die "No args";
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0+//?$_:$_ for@ARGV;
above 2147483647.
Ronald
P.S. Golfed further:
#!/usr/bin/perl -l
print"$_: ",(unpack"B*",pack N,$_)=~/(1.*)/for@ARGV?@ARGV:die"No args"
Olof Johansson
Nov 16, 2011, 10:14:57 AM11/16/11
to Ronald J Kimball, f...@perl.org
On 2011-11-16 09:57 -0500, Ronald J Kimball wrote:
> That should be s/0*//, otherwise you'll get the wrong result for numbers
> above 2147483647.
That should be s/^0*//, otherwise you'll get the wrong result for
> That should be s/0*//, otherwise you'll get the wrong result for numbers
> above 2147483647.
numbers above 2147483647.
Also:
printf"$_:%b\n",$_ for@ARGV
--
---------------------------------------------------------------
| Olof Johansson http://stdlib.se/ |
---------------------------------------------------------------
Ronald J Kimball
Nov 16, 2011, 11:57:50 AM11/16/11
to Olof Johansson, f...@perl.org
On Wed, Nov 16, 2011 at 04:14:57PM +0100, Olof Johansson wrote:
> On 2011-11-16 09:57 -0500, Ronald J Kimball wrote:
> > That should be s/0*//, otherwise you'll get the wrong result for numbers
> > above 2147483647.
>
> That should be s/^0*//, otherwise you'll get the wrong result for
> numbers above 2147483647.
No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
> On 2011-11-16 09:57 -0500, Ronald J Kimball wrote:
> > That should be s/0*//, otherwise you'll get the wrong result for numbers
> > above 2147483647.
>
> That should be s/^0*//, otherwise you'll get the wrong result for
> numbers above 2147483647.
the start of the string anyway.
Ronald
Olof Johansson
Nov 16, 2011, 1:49:14 PM11/16/11
to Ronald J Kimball, f...@perl.org
On 2011-11-16 11:57 -0500, Ronald J Kimball wrote:
> No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
> the start of the string anyway.
You're clearly an expert. I yield. Can you open a bug report with perl
> No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
> the start of the string anyway.
and getting this fixed?
Daniel Cutter
Nov 16, 2011, 1:49:12 PM11/16/11
to f...@perl.org
Hasn't anyone noticed that a decimal to binary or whatever conversion
isn't a JAPH by definition, alone because it does anything other than
output the JAPH? With a looser definition it must at least output "Just
another Perl hacker," perhaps on stderr
Daniel Cutter
@_=($_="aaceeehhjklnoprrrsttu", q<)411!**.&#)(#'&$!!"!!>);for$a(map{
'.'x(ord($_)-3*11)}$_[1]=~m,.,g){s,($a)(.),$1,,$b.=$2};$c.="(.{$_})"
for(4,7,4,6);@_=$b=~m,$c,;push@_,"\n";print map{ucfirst}map{$_.$"}@_
isn't a JAPH by definition, alone because it does anything other than
output the JAPH? With a looser definition it must at least output "Just
another Perl hacker," perhaps on stderr
Daniel Cutter
@_=($_="aaceeehhjklnoprrrsttu", q<)411!**.&#)(#'&$!!"!!>);for$a(map{
'.'x(ord($_)-3*11)}$_[1]=~m,.,g){s,($a)(.),$1,,$b.=$2};$c.="(.{$_})"
for(4,7,4,6);@_=$b=~m,$c,;push@_,"\n";print map{ucfirst}map{$_.$"}@_
Ronald J Kimball
Nov 16, 2011, 2:14:09 PM11/16/11
to Olof Johansson, f...@perl.org
On Wed, Nov 16, 2011 at 07:49:14PM +0100, Olof Johansson wrote:
> On 2011-11-16 11:57 -0500, Ronald J Kimball wrote:
> > No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
> > the start of the string anyway.
>
> You're clearly an expert. I yield. Can you open a bug report with perl
> and getting this fixed?
It's working as expected for me, so I'm not sure what needs to be fixed.
> On 2011-11-16 11:57 -0500, Ronald J Kimball wrote:
> > No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
> > the start of the string anyway.
>
> You're clearly an expert. I yield. Can you open a bug report with perl
> and getting this fixed?
% cat tmp.pl
#!perl -l
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0*//?$_:$_ for@_=@ARGV;
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/^0*//?$_:$_ for@_=@ARGV;
% perl tmp.pl 2147483648 3000000000
2147483648: 10000000000000000000000000000000
3000000000: 10110010110100000101111000000000
2147483648: 10000000000000000000000000000000
3000000000: 10110010110100000101111000000000
%
Are you seeing different behavior?
Ronald
Georg Moritz
Nov 16, 2011, 2:35:02 PM11/16/11
to f...@perl.org
From the keyboard of Ronald J Kimball [16.11.11,14:14]:
> On Wed, Nov 16, 2011 at 07:49:14PM +0100, Olof Johansson wrote:
>> On 2011-11-16 11:57 -0500, Ronald J Kimball wrote:
>>> No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
>>> the start of the string anyway.
rye
>> You're clearly an expert. I yield. Can you open a bug report with perl
>> and getting this fixed?
>
> It's working as expected for me, so I'm not sure what needs to be fixed.
well... the ternary ?: isn't necessary with s/0*// because s/0*// is
always successful, so
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0*//&&$_ for@ARGV;
two strokes ;-)
but as Olof pointed out,
printf"$_:%b\n",$_ for@ARGV
is less convoluted. Why do I forget about printf almost always?
'cause I'm not a C hacker, but a heck parler, I guess ;-)
for some JAPH, see my signature. Latin-1 only, though
cheers,
0--gg-
> % cat tmp.pl
> #!perl -l
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0*//?$_:$_ for@_=@ARGV;
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/^0*//?$_:$_ for@_=@ARGV;
> % perl tmp.pl 2147483648 3000000000
> 2147483648: 10000000000000000000000000000000
> 3000000000: 10110010110100000101111000000000
> 2147483648: 10000000000000000000000000000000
> 3000000000: 10110010110100000101111000000000
> %
>
> Are you seeing different behavior?
>
> Ronald
>
> On Wed, Nov 16, 2011 at 07:49:14PM +0100, Olof Johansson wrote:
>> On 2011-11-16 11:57 -0500, Ronald J Kimball wrote:
>>> No, you are wrong. s/0*// is sufficient, because /0*/ will always match at
>>> the start of the string anyway.
>> You're clearly an expert. I yield. Can you open a bug report with perl
>> and getting this fixed?
>
> It's working as expected for me, so I'm not sure what needs to be fixed.
always successful, so
print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0*//&&$_ for@ARGV;
two strokes ;-)
but as Olof pointed out,
printf"$_:%b\n",$_ for@ARGV
'cause I'm not a C hacker, but a heck parler, I guess ;-)
for some JAPH, see my signature. Latin-1 only, though
cheers,
0--gg-
> % cat tmp.pl
> #!perl -l
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/0*//?$_:$_ for@_=@ARGV;
> print"$_: ",($_=unpack"B*",pack"N",$_)=~s/^0*//?$_:$_ for@_=@ARGV;
> % perl tmp.pl 2147483648 3000000000
> 2147483648: 10000000000000000000000000000000
> 3000000000: 10110010110100000101111000000000
> 2147483648: 10000000000000000000000000000000
> 3000000000: 10110010110100000101111000000000
> %
>
> Are you seeing different behavior?
>
> Ronald
>
Olof Johansson
Nov 16, 2011, 3:34:58 PM11/16/11
to Ronald J Kimball, f...@perl.org
On 2011-11-16 14:14 -0500, Ronald J Kimball wrote:
> It's working as expected for me, so I'm not sure what needs to be fixed.
Hum, now I'm ashamed. Sorry. But why is that not greedy? (I got other
> It's working as expected for me, so I'm not sure what needs to be fixed.
results earlier, but I can't reproduce now so I have probably made an
error in my earlier attempts.)
John Douglas Porter
Nov 16, 2011, 3:40:57 PM11/16/11
to Ronald J Kimball, Olof Johansson, f...@perl.org
Remember, the *first* match wins, even if it's shorter
than a possible later match.
ISTR that some have argued that's a bug.
Well, too bad. It's too late. :-)
--
john "many jars" porter
than a possible later match.
ISTR that some have argued that's a bug.
Well, too bad. It's too late. :-)
--
john "many jars" porter
Ronald J Kimball
Nov 16, 2011, 3:46:52 PM11/16/11
to Olof Johansson, f...@perl.org
On Wed, Nov 16, 2011 at 09:34:58PM +0100, Olof Johansson wrote:
> On 2011-11-16 14:14 -0500, Ronald J Kimball wrote:
> > It's working as expected for me, so I'm not sure what needs to be fixed.
>
> Hum, now I'm ashamed. Sorry. But why is that not greedy? (I got other
> results earlier, but I can't reproduce now so I have probably made an
> error in my earlier attempts.)
It is greedy, but the important thing to remember is that the regular
> On 2011-11-16 14:14 -0500, Ronald J Kimball wrote:
> > It's working as expected for me, so I'm not sure what needs to be fixed.
>
> Hum, now I'm ashamed. Sorry. But why is that not greedy? (I got other
> results earlier, but I can't reproduce now so I have probably made an
> error in my earlier attempts.)
expression engine will find the longest *leftmost* match.
For example, "00111000" =~ /0*/ will match "(00)111000". It won't match
"(0)0111000", because, although it's a leftmost match, it's not the longest
leftmost match. And it won't match "00111(000)" because that's not a
leftmost match, regardless of the length.
Ronald
Aristotle Pagaltzis
Nov 18, 2011, 3:41:43 PM11/18/11
to f...@perl.org
* Ronald J Kimball <rjk-pe...@tamias.net> [2011-11-16 21:50]:
location, and will make the match as long as possible at that location.
> It is greedy, but the important thing to remember is that the regular
> expression engine will find the longest *leftmost* match.
To put that a third way: the engine will match at the first possible
> expression engine will find the longest *leftmost* match.
location, and will make the match as long as possible at that location.
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