Given

sub Tie::TIESCALAR { return bless \$_[1] => $class }
sub Tie::FETCH { return ${$_[0]} }
sub Tie::STORE { $_[0] = "I'm not tied any more" }
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UPDATE: Oops, $class should be $_[0].

Yes, providing quite broken 'tie' implementations can make any code blow up. q-:

Fixing your code so that it works and also demonstrates that things don't stop being tied shows that my code works fine with such things:

#!/usr/bin/perl -w
use strict;

# my $x = 1;
tie my $x, 'Tie', 1;
my $f = sub { $x };
print $f->(), "\n"; # => 1
{
    my $scope= tempSet( \$x, 2 );
    print $f->(), "\n"; # => 2
}
print $f->(), "\n"; # => 1

sub TempSet::DESTROY { shift(@_)->() }
sub tempSet {
    my( $sv, $new )= @_;
    my $prior= $$sv;
    my $restore= bless
        sub { $$sv= $prior },
        'TempSet';
    $$sv= $new;
    return $restore;
}

sub Tie::TIESCALAR { return bless \$_[1] => $_[0] }
sub Tie::FETCH { return "${$_[0]} is still tied at line " . (caller())
+[2] }
sub Tie::STORE { ( ${$_[0]} = $_[1] ) =~ s/ is still tied at line \d+$
+// }
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which produces

1 is still tied at line 7
2 is still tied at line 10
1 is still tied at line 12
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(Updated code to make tied-ness more apparent.)

(Yes, in constrast, local is implemented in a way that it temporarily hides the tied nature of the localized scalar.)

- tye        

Fixing your code so that it works and also demonstrates that things don't stop being tied shows that my code works fine with such things …

I agree that my code isn't a very good implementation of a tied variable, but it is what I meant (UPDATE 3: except for the silly out-of-place $class), and fixing it breaks the example. :-) (Your code doesn't quite fix it, because you try to assign to a read-only variable. Something like sub Tie::TIESCALAR { return bless \( my $o = $_[1] ) => $_[0] } would be better, I think.)

As a less offensive example, if I replace your STORE by

sub Tie::STORE {
   ${$_[0]} = $_[1] unless defined ${$_[0]}
}
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then it's clear that the localisation isn't really protecting anything in the inner scope from changing:

tie my $x, 'Tie';
my $f = sub { $x };
{
    my $scope = tempSet(\$x, 2);
    print $f->(), "\n"; # => 2 is still tied
}
$x = 1;
print $f->(), "\n"; # => 2 is still tied
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Your comment

Yes, in contrast, local is implemented in a way that it temporarily hides the tied nature of the localized scalar.

hits the nail on your head—I want precisely something that has the same effect for lexicals, of setting aside their previous value, whatever that value is, and restoring it undisturbed afterwards. (In particular, $x should not be tied in the inner scope, even if it is outside.)

UPDATE 1: Actually, testing with tie our $x, 'Tie' and an interior local $x seems to show that localisation of globals doesn't behave the way I want, either! (See Re^6: Local for lexicals (untie) for a simpler example.)
UPDATE 2: Changed if defined test to unless defined.

#!perl
use strict;
use warnings;

my $x = 1;
sub lambda
{
    my $fn = pop;
    my @argrefs = \(@_);
    sub {
      my @scopes= map { tempSet( $argrefs[$_], $_[$_] ) } 0..$#_;
      &$fn;
    }
}

my $y = 'Yval';
my $closure = lambda $x, $y => sub { "$x and $y" };

print $closure->($_, rand(3)), "(X=$x, Y=$y)\n" for (qw(a bitty china 
+doll));

print "$x and $y\n"; # => 1

sub TempSet::DESTROY { shift(@_)->() }
sub tempSet {
    my( $sv, $new )= @_;
    my $prior= $$sv;
    my $restore= bless
        sub { $$sv= $prior },
        'TempSet';
    $$sv= $new;
    return $restore;
}
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