perlop is clear that and has higher precedence than or and this can be checked with B::Deparse and -p option

$_ % 2 or ( $has_even = 1 and last )

Like others already implied does $has_even = 1 and last not make much sense, since the assignment at LHS is always true.

Furthermore is this a quite pathological case, because the part with the code-branch with higher precedence is never executed.

compare

  DB<215> p 0 or 1 and 0
0
  DB<216> p (0 or 1) and 0
1
  DB<217>
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update

  DB<233> p 0 or 1 and die "never"
0
  DB<234> p 0 or (1 and die "never")
0
  DB<235> p (0 or 1) and die "never"
1never at (eval 265)[c:/Perl_524/lib/perl5db.pl:737] line 2.

  DB<236>
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