In the following code, what does \Q do to make things work as expected and why do the other attempts fail at working as expected?

UPDATE: figured out the explicit versions

UPDATE: Got the \Q part after reading the link given by Corion: \Q makes $x into $x='E\\:\\\\th\\\\foo' 

#!c:/opt/perl/bin/perl
use warnings;
use strict;

my $t='E:\\th\\foo\\bl\\be';
my $x='E:\\th\\foo';

print "t:$t, x:$x\n";

# m// should show that $t starts with $x

$t =~ m,^$x,   and print "matched\n"           or print "didn't match\
+n";
$t =~ m,^\Q$x, and print "matched with Q\n"    or print "didn't match 
+with Q\n";
$t =~ m,^\$x,  and print "matched with 1 \\\n" or print "didn't match 
+with 1 \\\n";
$t =~ m,^\\$x, and print "matched with 2 \\\n" or print "didn't match 
+with 2 \\\n";
# tried with 3 and 4 \ too

# being explicit on right (no quotes)
$t =~ m,E:\\th\\foo, and print "yes (explicit)\n" or print "no (explic
+it)\n";
$t =~ m,E\:\\th\\foo, and print "yes (1 \\ :)\n" or print "no (1 \\ :)
+\n";
$t =~ m,E\\:\\th\\foo, and print "yes (2 \\ :)\n" or print "no (2 \\ :
+)\n";



__END__
t:E:\th\foo\bl\be, x:E:\th\foo
didn't match
matched with Q
didn't match with 1 \
didn't match with 2 \
yes (explicit)
yes (1 \ :)
no (2 \ :)

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In reply to Working of \Q in m// by sg

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