I don't think that it really is.

Why not? How does "treated like any vulgar scalar" differ from "no longer treated specially"? (I said phrase, not sentence)

A commutativity breakage means that there are some circumstances in which the smart match depends on order, not that the smart match is always completely determined by the right-hand member.

I know. I never said otherwise.

In reply to Re^4: Perl 5.11.0 now available by ikegami
in thread Perl 5.11.0 now available by Anonymous Monk

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