My dense skull isn't penetrating this issue. The original poster wrote 
my $x;  # one statement
$x = $y if {condition;}   # two statement

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Both when I posted and now, the OP reads: my $x = $y if ( <some condition> ); Are you looking at someone else's reply?

With the my $x separate and unconditional, the warning in perlsyn doesn't apply.

The output from your strict version doesn't make sense to me, offhand.

In reply to Re^3: "if" in a my declaration statement causes problem by ysth
in thread "if" in a my declaration statement causes problem by ganeshk

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