use strict;
use warnings;

use constant DLE => 0x10;

my $dle = sprintf ("%02X", DLE);

print "before\n";
print "string is: [\x{$dle}]\n";

__output__

Illegal hexadecimal digit '$' ignored at pmsopw_294950.pl line 9.
before
string is: []
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use strict;
use warnings;

my $msg = pack "CCCCC", 0x31, 0x32, 0x10, 0x10, 0x33;
my $c   = pack "C", 0x10;

$msg =~ s/$c$c/$c/g;

printf ("%02x " x length($msg) . "\n", unpack ("C*", $msg));

__output__

31 32 10 33
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#!/usr/bin/perl -w

use strict;
use constant DLE=>0x10;
use vars qw($DLE_C);
$DLE_C = chr(DLE);

my $msg = pack ("CCCCC", 0x31, 0x32, 0x10, 0x10, 0x33);
# match $dle's expanded, replacement is not expanded ?
$msg =~ s/$DLE_C$DLE_C/$DLE_C/g;
printf ("%02x " x length($msg) . "\n", unpack ("C*", $msg));
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Another way is to use capturing inside the RE, then just replace with the captured part:

$msg =~ s/(\x{$dle})\x{$dle}/$1/g;
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As for why it doesn't work the way you expect, Abigail-II seems to have already given a better answer than I could.

I like the chr() approach; however, I was concerned about getting tangled up in multibyte unicode problems. It *looks* like if I use chr(???) with a ??? <= 255 I will always get the single byte I am looking for (i.e. not translated to/from some type of unicode symbol set). Correct ?
With respect to your second suggestion:
1. Yes, I also had this thought. I was originally trying to run with .../go; (for efficiency) and I was concerned that the captured string would not be inserted without recompilation. I was probably mistaken ... :)
2. The fact that this works implies the the initial \x{$dle} IS interpolated and the replacement one IS NOT. Abigail-II's explanation did not cover why one is interpolated and the other is not. Seems unduly inconsistant - even for perl :)
Thanks to all ! (Is it customary to reply with "thanks" (only), or is that considered unnecessary babble ?) Thanks, Scott.

It *looks* like if I use chr(???) with a ??? <= 255 I will always get the single byte I am looking for (i.e. not translated to/from some type of unicode symbol set). Correct ?

Well... in a way... yes. But you're overlooking one thing: if Perl concatenates a UTF8 string with a Latin-1 string (at least, that's the only way to think about it that makes sense), Perl will convert the Latin-1 string to UTF-8. Let me show you with an example:

($\, $,) = ("\n", " "); # set up output mode
$string = "A" . chr(180) . "B";  # Latin-1
print unpack "C*", $string;
$string .= chr(367); # UTF-8
print unpack "C*", $string;
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Output:

65 180 66
65 194 180 66 197 175

As you can see, the original chr(180), between chr(65) ("A") and chr(66) ("B") is converted to UTF-8, rsulting in two bytes.

So, if you want UTF-8, all you have to do is insert the characters into a UTF-8 string, or concatenate it with a UTF-8 string. That may even be a zero-length string, asq returned by pack "U0":

($\, $,) = ("\n", " "); # set up output mode
$string = "A" . chr(180) . "B";  # Latin-1
print unpack "C*", $string;
$string .= pack "U0"; # zero length, UTF-8
print unpack "C*", $string;
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Result:

65 180 66
65 194 180 66

p.s. This was tested with perl 5.6.1. on Windows. Not that it matters much — it shouldn't, except that you need at least perl 5.6.