Bit Shifting ?

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

I know the uber Perlies will unravel this one without
breaking a sweat. Perhaps a the newer Perlites will take
a crack.

Why does this:

my ($perms, $type) = @ARGV;
print "Other Test: ", ($perms & $type ) ,"\n";
print "Group Test: ", ($perms & ($type << 3 )),"\n";
print "Owner Test: ", ($perms & ($type << 4)),"\n";
# with command line params: 63,1;
[download]

Produce this:

Other Test: 0
Group Test: 8
Owner Test: 16

And this:

my ($perms, $type) = @ARGV;
print "Other Test: ", ($perms & ($type << 0) ,"\n";
print "Group Test: ", ($perms & ($type << 3 )),"\n";
print "Owner Test: ", ($perms & ($type << 4)),"\n";
# with command line params: 63,1;
[download]

Produce this:

Other Test: 1
Group Test: 8
Owner Test: 16

Well snap my deerskin thong! I just figured her out,
but shoot I think I'll post her anyways.

mitd-Made in the Dark
I've always been astonished by the absurd turns
rivers have to make to flow under every bridge.

Comment on Bit Shifting ? Select or Download Code

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Re: Bit Shifting ?
by MarkM (Curate) on Apr 06, 2003 at 03:21 UTC

@ARGV is an array of strings. Bitwise OR or AND of strings in Perl results in a byte-by-byte OR or AND. Compare:

$ perl -e "print 1 & 63, qq(\n)"
1
$ perl -e "print q(1) & q(63), qq(\n)"
0
[download]

As an integer bitwise AND, the result is 1. As a string bitwise AND, the result is 0. Why? As a string, ord("1") & ord("6") yields ord("0"). To try something cool, try:

$ perl -e "print q(a) & q(ab), qq(\n)"
a
[download]

Similarily, ord("a") & ord("a") yields ord("a"). (The characters on the end don't count for bitwise AND since the strings have a different length, and Perl doesn't explicitly include the "\0" characters at the end)

Any time you have a scalar that may be in string context that you need to manipulate as a number, consider using a common expression such as '0+$scalar' to force it to have a numeric representation. @ARGV appears to be just such a case.

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