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Re^2: ternary operator

by japhy (Canon)
on Jul 17, 2006 at 22:55 UTC ( #561881=note: print w/replies, xml ) Need Help??


in reply to Re: ternary operator
in thread ternary operator

No bug. The statement parses as ($test ne 'c' ? ($OPER = 's') : $OPER) = 'c';

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

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Re^3: ternary operator
by GrandFather (Saint) on Jul 17, 2006 at 23:30 UTC

    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. However no errors and no warnings are generated for either condition.

    Using ?: to select a variable to be assigned to I'm happy with:

    test ? $p1 : $p2 = value;

    but the half assignement variant implied by the precedence (and made explicit by the brackets shown in your reply) is just bizare.


    DWIM is Perl's answer to Gödel

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

      print "Just another Perl ${\(trickster and hacker)},"
      The Sidhekin proves Sidhe did it!

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