bit-wise operation

toro_the_third has asked for the wisdom of the Perl Monks concerning the following question:

Hello, I'm new here and let me apologize for any inconvenience regarding this post.
so, my question is that I'm trying to make a script that reads 2　arguments and returns result of bit-wise operation AND.

#!/usr/bin/perl
use strict;
use warning;

my $arg00=shift ( @ARGV );
my $arg01=shift ( @ARGV );

#printf $arg00+$arg01."\n";
printf sprintf( $arg00&$arg01 )."\n";
[download]

when I run test.pl 33 224, it returns result as 22.
if I run printf sprintf( 33&224 )."\n"; it returns 32. also if I comment out the # before the printf, it also returns 32.
could any one give me an advice about this difference?
regards,
toro

Comment on bit-wise operation Download Code

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Re: bit-wise operation (+) by tye (Sage) on Feb 28, 2012 at 06:42 UTC
Ugh. The documentation on `&` in perlop is split into two, one half under "Bitwise And" and the other half under "Bitwise String Operators" and neither half even hints at how `&` can do both of these different operations. So, I guess it is somewhere else in the documentation (or perhaps even only in earlier versions of the documentation) where I have read that `&` does bit-wise integer 'and' if either of the two arguments have been recently used as a number. However, if both arguments have only ever been used as strings, then bit-wise string 'and' is done. `@ARGV` contains the string arguments given on the command line. This (the bit-wise operators, `&` and `\|`) is one of the quite rare cases where this kind of tricky detail matters. I'd probably change your code to use: `my $arg00 = 0 + shift @ARGV;` [download] to force `$arg00` to be treated as a number. Or you can force it right next to the operator: `(0+$arg00) & $arg01` [download] - tye	[reply] [d/l] [select]
Re^2: bit-wise operation (+) by toro_the_third (Novice) on Feb 28, 2012 at 07:03 UTC
Thank you very much for your great help. I have fixed the variable declaration part and worked fine. Now, I understand why it worked when I put printf $arg00+$arg01."\n"; between this code. Thanks again for your help, I have been suffering for days about this. thanks, toro `[root@local]#cat test.pl #!/usr/bin/perl use strict; use warning; my $arg00=0+shift ( @ARGV ); my $arg01=0+shift ( @ARGV ); #printf $arg00+$arg01."\n"; printf( "%d", $arg00 & $arg01 ); [root@local]#./test.pl 33 224 <BR>32[root@local]#` [download]	[reply] [d/l]
Re: bit-wise operation by NetWallah (Canon) on Feb 28, 2012 at 06:13 UTC
Both printf and sprintf take TWO arguments: `printf FORMAT, LIST` [download] You supplied only one, so it prints the content of what you supplied. Try using "%X" as a format, to print hex, or review the sprintf documentation for more options. I do not have an explanation for your "32" result, but I can reproduce it. Perhaps a wiser monk .... “PHP is a minor evil perpetrated and created by incompetent amateurs, whereas Perl is a great and insidious evil perpetrated by skilled but perverted professionals.” ― Jon Ribbens	[reply] [d/l]
Re^2: bit-wise operation by Anonymous Monk on Feb 28, 2012 at 06:35 UTC
Hi, thank you for the prompt reply. I have fixed the printf part. but it still wouldn't work correct. `[root@local]#cat test.pl #!/usr/bin/perl use strict; use warning; my $arg00=shift ( @ARGV ); my $arg01=shift ( @ARGV ); #printf $arg00+$arg01."\n"; printf( "%d", $arg00 & $arg01 ); [root@local]#./test.pl 33 224 <BR>22[root@local]#` [download] It should return 32 but returns 22. I cant figure out why this happens... regards, toro	[reply] [d/l]