The first one is interpreted as:
print $_ . ("\n" && next) if /no_proc/;

The  . operator has higher precedence than the  && operator so it is actually interpreted as: 

$ perl -MO=Deparse,-p -e' print $_ . "\n" && next if /no_proc/; '
(/no_proc/ and print((($_ . "\n") && next)));
-e syntax OK

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In reply to Re^3: if modifier of a print statement by jwkrahn
in thread if modifier of a print statement by si_lence

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