in reply to Why won't $' work in a trinary operation?

This gotcha is explained in the standard documentation for the conditional operator in perlop:
Because this operator produces an assignable result, using assignments without parentheses will get you in trouble. For example, this: $a % 2 ? $a += 10 : $a += 2 Really means this: (($a % 2) ? ($a += 10) : $a) += 2 Rather than this: ($a % 2) ? ($a += 10) : ($a += 2)