which to me looks like: $OPER = 's' = 'c'; when $test ne 'c' is true. And that should fail with a "Can't modify constant item in scalar assignment" error.

You're forgetting the precedence -- or should I say associativity? -- no, I think precedence is correct here. It is actually like ($OPER = 's') = 'c', which, while admittedly bizarre, is perfectly legal, since $OPER = 's' evaluates to an lvalue $OPER. Observe the difference:

sidhekin@blackbox:~$ perl -le '($c = 1) = 2; print $c'
2
sidhekin@blackbox:~$ perl -le '$c = 1 = 2; print $c'
Can't modify constant item in scalar assignment at -e line 1, near "2;
+"
Execution of -e aborted due to compilation errors.
sidhekin@blackbox:~$
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