Re: ternary operator

There may be a Perl bug here. Because of operator precedence I'd have expected that the trinary experession shouldn't compile. The assignment operator has lower percedence than the trinary operator.

However the immediate answer to your problem is to fix the precedence issue. Here are two fixes:

1/ $test ne 'c' ? ($OPER='s') : ($OPER='c');
2/ $OPER = $test ne 'c' ? 's' : 'c';
[download]

DWIM is Perl's answer to Gödel

Comment on Re: ternary operator Download Code

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Re^2: ternary operator by japhy (Canon) on Jul 17, 2006 at 22:55 UTC
No bug. The statement parses as `($test ne 'c' ? ($OPER = 's') : $OPER) = 'c';` Jeff `japhy` Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and `perl` hacker How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart	[reply] [d/l]
Re^3: ternary operator by GrandFather (Saint) on Jul 17, 2006 at 23:30 UTC
which to me looks like: `$OPER = 's' = 'c';` [download] 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 : $p2 = value;` [download] but the half assignement variant implied by the precedence (and made explicit by the brackets shown in your reply) is just bizare. DWIM is Perl's answer to Gödel	[reply] [d/l] [select]
Re^4: ternary operator by Sidhekin (Priest) on Jul 17, 2006 at 23:45 UTC
which to me looks like: `$OPER = 's' = 'c';` when $test ne 'c' is true. And that should fail with a "Can't modify constant item in scalar assignment" error. You're forgetting the precedence -- or should I say associativity? -- no, I think precedence is correct here. It is actually like `($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 $c' Can't modify constant item in scalar assignment at -e line 1, near "2; +" Execution of -e aborted due to compilation errors. sidhekin@blackbox:~$` [download] `print "Just another Perl ${\(trickster and hacker)},"` The Sidhekin proves Sidhe did it!	[reply] [d/l] [select]


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