$dereferencer = { .<foo>[1]<bar><baz>[3] };

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I knew that "unary" dot acts implicitly on $_, but appearently your code implies that in code written as a bare block $_ is an alias to the first positional argument, and that I didn't know. Actually, without knowing, I would have written the above amongst other possibilities, like:

$dereferencer = -> $d { $d.<foo>[1]<bar><baz>[3] };

[download]

In reply to Re^4: [Perl 6] Any provision for a "dereferencing object"? by blazar
in thread [Perl 6] Any provision for a "dereferencing object"? by blazar

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