[perl #66782] [BUG] calling .=subst leaves $/ undefined in the closure in the second parameter in Rakudo

[Carl Mäsak]

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<masak> rakudo: $_ = 1; .say; .=subst(/(.) $0*/, { $/.chars ~ $0 }, :g); .say
<p6eval> rakudo 1b06df: OUTPUT«1␤Use of uninitialized value␤Use of
uninitialized value␤0␤»
<masak> I expected that one to say '11' the second time.
<masak> what am I missing?
<pmichaud> I'm not sure.
<masak> rakudo: $_ = "foo"; say .subst(/o/, { ~$/ }, :g)
<p6eval> rakudo 1b06df: OUTPUT«foo␤»
<masak> rakudo: $_ = "foo"; say .=subst(/o/, { ~$/ }, :g)
<p6eval> rakudo 1b06df: OUTPUT«Use of uninitialized value␤Use of
uninitialized value␤f␤»
* masak submits rakudobug
<TimToady> {} has its own $/ maybe?
<masak> some strange interaction between .= and $/
<masak> TimToady: that doesn't explain why .subst works above.
<pmichaud> it might be an odd interaction with $_ .  Or perhaps with
the way .= has been implemented.