perldoc perlsub says: 
The declared variable is not introduced (is not visible)
until after the current statement.  Thus,

         my $x = $x;

can be used to initialize a new $x with the value of the old$x, [...]

[download]
So perl tries to find another $epoch.

HTH

In reply to Re: my $variable with &&do{} doesn't compile by Skeeve
in thread my $variable with &&do{} doesn't compile by Biker

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