in reply to $_ as default printf arg fails

I agree that the doc seems to be wrong. The doc even implies that the following should do something, but I still get a warning:
printf;

The current printf (16.2) has the same information. This may be worthy of a perlbug.

UPDATE: This doc first appeared on 14.2 (it wasn't there on 14.0, or before), but I don't see a change to printf behavior in perl5141delta or perl5142delta.

It seems the warning is triggered if $_ contains %

$_ = "%d 42"; printf;

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Re^2: $_ as default printf arg fails
by BillKSmith (Monsignor) on Mar 21, 2013 at 14:20 UTC

    I believe that the document is correct. The default $_ is treated as a FORMAT. The warning is triggered (or not) the same as it is for an explicit FORMAT. Subtle isn't it?

    Bill