I believe this is because the \x{...} is evaluated at compile-time, and the characters between braces aren't interpolated e.g
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: []
A simpler solution would be just to continue using pack e.g
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

HTH

_________
broquaint


In reply to Re: Why is variable interpolation suppressed in \x{$xxx} replacement ? by broquaint
in thread Why is variable interpolation suppressed in \x{$xxx} replacement ? by Anonymous Monk

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