I don't understand. Which sub I don't call? The sub which creates the guard? Of course it is called! It is stored in $code and it is called immediately after do. Your example is something very different.

The only purpose of do is to create a variable which is referenced only from the anonymous sub.

my $x;
my $code = do {
    my $g;
    sub { $g = Scope::Guard->new( sub { warn "destroyed";  $x ; 
} ) };
};
$code->();
undef $code;
warn "end";

[download]

I also don't understand the rest of your comment. The problem is not that $g references something, but that it is referenced from somewhere. But what keeps the reference to $g?

My example is the very simplification of real scenario I came across in asynchronous programming. I have two callbacks sharing a resource (handle). The resource is stored in lexical variable visible only in these two subs. First callback is called, creates the resource ($code->()) and goes out of scope. Second callback is scheduled to run. Than the process is cancelled and the second callback is cancelled from queue and goes out of scope (undef $code - here I simplified to one callback only). I expect the resource to be released then.

In reply to Re^2: Reference to guard not released by roman
in thread Reference to guard not released by roman

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