Problem: Given a binary string of a principally indeterminate (max 8
length bytes), turn that into a hex dump with 16 bytes per line (and
possibly less on the last line.

Solution:

---------
sub format_bin
{ use bytes;
my @lines;
local *_ = \$_[0];

push(@lines, join(' ', unpack('(H2)*', $1)))
while /\G(.{1,16})/gs;

return @lines;
} ----------

I'm especially happy that there's no need to special-case the last line
as the /\G(.{1,16})/gs will match whatever is remains after the last
complete 16-character-group.

On Thu, 06 May 2021 18:55:38 +0100, Rainer Weikusat wrote:

> local *_ = \$_[0];

What witchcraft is that? I don't even know where to start looking it up.

--
Henry Law n e w s @ l a w s h o u s e . o r g
Manchester, England

Henry Law <news@lawshouse.org> writes:
> On Thu, 06 May 2021 18:55:38 +0100, Rainer Weikusat wrote:
>
>> local *_ = \$_[0];
>
> What witchcraft is that? I don't even know where to start looking it up.

It's documented in the "Symbol Tables" section of perlmod: *_ is the
typeglob whose scalar slot is $_. $_[0] is the scalar passed as first
argument (ie, at the implementation level, it's a SV *). Assigning a
reference to that to a glob causes the scalar slot of this glob to refer
to the scalar the reference came from: Afterwards (while the local is in
scope) $_ 'means' 'the scalar passed as first argument': It now has a
name but its contents weren't copied.

Rainer Weikusat <rweikusat@talktalk.net> writes:

# [1]
local *_ = \$_[0];

> ... to the scalar the reference came from: Afterwards (while the local is in
> scope) $_ 'means' 'the scalar passed as first argument': It now has a
> name but its contents weren't copied.

I wonder how it compares to this alternative:

# [2]
local $_ = $_[0];

I guess [1] is aliasing and [2] is copying, although I'm not 100%
confident about this (nor the exact meaning of aliasing.)