Bit testing

I'm almost embarrassed to ask this, but I can't figure out a simple way =
to construct a switch ('given') statement where the 'when' clauses =
involve bit-testing.

Here's the only way I've figured out to build a switch statement that =
does the trick. It seems unusually wordy, which makes me think there =
must be a simpler way to test for certain bit combinations. Any =
suggestions?

Thanks,
Chap

#!/usr/bin/perl =
=
=20

use strict;
use warnings;
use feature ":5.10";

# Here are masks for various bit combos of interest: =
=
=20

my $one_three =3D 0b00001010; # bits 1 and 3 (counting from 0, =
right to left) =
=20
my $zero_four =3D 0b00010001; # bits 0 and 4 =
=
=20
my $five =3D 0b00100000; # bit 5 =
=
=20

# Here we will test several bit fields for bit combos of interest: =
=
=20

for my $flags ( 0b10111010, 0b10111000, 0b10010010) {

my $asbits =3D sprintf("0b%08b", $flags); # prepare bits for =
pretty-printing =
=20

given ( $flags ) {
when ( ($_ & $one_three) == $one_three ) { # bits one and =
three are on =
=20
say "$asbits has bits 1 and 3";
}
when ( ($_ & $zero_four) == $zero_four ) { # bits zero and =
four are on =
=20
say "$asbits has bits 0 and 4";
}
when ( ($_ & $five) == $five ) { # bit five is on =
=
=20
say "$asbits has bit 5";
}
default {
say "$asbits has no interesting bit patterns.";
}
}
}



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Re: Bit testing

>>>>> "CH" == Chap Harrison writes:

CH> I'm almost embarrassed to ask this, but I can't figure out a
CH> simple way to construct a switch ('given') statement where the
CH> 'when' clauses involve bit-testing. Here's the only way I've
CH> figured out to build a switch statement that does the trick. It
CH> seems unusually wordy, which makes me think there must be a
CH> simpler way to test for certain bit combinations. Any
CH> suggestions?

CH> when ( ($_ & $one_three) == $one_three ) { # bits one and
CH> three are on

don't put comments on the same line as code. it can make it line wrap
like above.

anyhow, some boolean hacking does the trick. this seems to work.


when ( !(~$_ & $one_three) )

you invert the flag bits and then and against the test mask. if that is all
0's (tested with !) then your bits were set.

uri

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Re: Bit testing

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Re: Bit testing

Hi Charles,

On Sunday 14 November 2010 01:47:36 C.DeRykus wrote:
> On Nov 11, 11:27 pm, c...@pobox.com (Chap Harrison) wrote:
> Not lots shorter but you could use a closure to hide
> the calculation:
>
> my $mask;
> for my $flags ( ... ) {
> $mask = sub { return ($flags & $_[0]) == $_[0] }
> unless $mask;
> given( $flags ) {
> when ( $mask->($one_and_three) ) { ... }
> when ( $mask->($zero_and_four) ) { ... }
> ...
> }
> }
>

This won't work properly because the closure traps the initial value of
"$flags". For example:

[code]
#!/usr/bin/perl

use strict;
use warnings;

my $closure;

foreach my $name (qw(Sophie Jack Charles Dan Rachel))
{
$closure = sub { print "Hello $name!\n" ; } unless $closure;

$closure->();
}
[/code]

This prints "Hello Sophie!" five times. Either redeclare the closure on every
iteration, or declare it once while using a more outer lexical variable.

Regards,

Shlomi Fish

> --
> Charles DeRykus

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Re: Bit testing

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Re: Bit testing

On Nov 14, 2010, at 4:36 AM, C.DeRykus wrote:

> And now it seems a little bit
> inelegant to redefine the closure each time through
> the loop.
>=20
>=20
> for my $flags ( ... ) {
> my $mask =3D sub { return ($flags & $_[0]) == $_[0] };
> given( $flags ) {
> when ( $mask->($one_and_three) ) { ... }
> when ( $mask->($zero_and_four) ) { ... }
> ...
> }
> ...

First, thanks for (($flags & $mask) == $mask).=20

What I eventually did was just call a subroutine on($flags, $mask) to do =
the calculation and determine whether all the specified bits were on. =
Seems simpler than the closure technique, even if having to repeatedly =
write '$flags' is less than ideal. =20

However, now I'm determined to get an understanding of closures, and I =
see another thread on this list that provides several links to help in =
that regard.

Thanks, one and all!

Regards,
Chap


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Re: Bit testing

On Nov 14, 1:11=A0am, shlo...@iglu.org.il (Shlomi Fish) wrote:
> Hi Charles,
>
> On Sunday 14 November 2010 01:47:36 C.DeRykus wrote:
>
> > On Nov 11, 11:27 pm, c...@pobox.com (Chap Harrison) wrote:
> > Not lots shorter but you could use a closure to hide
> > the calculation:
>
> > my $mask;
> > for my $flags ( ... ) {
> > =A0 =A0 =A0$mask =3D sub { return ($flags & $_[0]) == $_[0] }
> > =A0 =A0 =A0 =A0 =A0 =A0 unless $mask;
> > =A0 =A0 =A0given( $flags ) {
> > =A0 =A0 =A0 =A0 =A0 =A0 when ( $mask->($one_and_three) =A0) { ... }
> > =A0 =A0 =A0 =A0 =A0 =A0 when ( $mask->($zero_and_four) =A0 ) { ... }
> > =A0 =A0 =A0 =A0 =A0 =A0 ...
> > =A0 =A0 =A0}
> > }
>
> This won't work properly because the closure traps the initial value of
> "$flags". For example:
>
> [code]
> #!/usr/bin/perl
>
> use strict;
> use warnings;
>
> my $closure;
>
> foreach my $name (qw(Sophie Jack Charles Dan Rachel))
> {
> =A0 =A0 $closure =3D sub { print "Hello $name!\n" ; } unless $closure;
>
> =A0 =A0 $closure->();}
>
> [/code]
>
> This prints "Hello Sophie!" five times. Either redeclare the closure on e=
very
> iteration, or declare it once while using a more outer lexical variable.

Right... or simply get rid of the statement qualifier
'unless $closure' which'll work too and is what you'll
have to do in any case.
(technically, you could declare 'my $closure' both
in/outside the loop leaving the statement qualifier
as is but that's horrible )

You could just 'my $closure' solely inside the loop
too.

Declaring once outside the loop with an outer lexical
seems less satisfactory since it loosens the 'tightest
lexical scope' best practice.

--
Charles DeRykus


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