Odd Perl conditional operator behavior

Question

I'm doing some work in Perl and I ran across an odd result using the conditional operator.

The code in question:

($foo eq "blah") ? @x = @somearray : @y = ("another","array");

Trying to compile this code results in the error "Assignment to both a list and a scalar at XXX line YY, near ');'". In trying to pinpoint the source of the error I've written this using a couple different ways of representing an array in Perl and they all return with the same error. Now at first I thought it was just some dumb obvious mistake with the assignment statements, but just to satisfy my curiosity I rewrote the statement in a more verbose way:

if($foo eq "blah") {
    @x = @somearray;
} else {
    @y = ("another","array");
}

That version of the code compiled perfectly fine.

Is there some fine distinction between how the conditional operator works and a basic if-else statement works that I'm missing here? I always understood the conditional operator to be just a short-hand version of the second statement. If there isn't a functional difference between the two, why would Perl object to the first statement, but not the second?

Answer 1

$ perl -MO=Deparse -e'($foo eq "blah") ? @x = @somearray : @y = ("another","array");'
Assignment to both a list and a scalar at -e line 1, near ");"
-e had compilation errors.
$foo eq 'blah' ? (@x = @somearray) : @y = ('another', 'array');
$ perl -MO=Deparse -e'($foo eq "blah") ? @x = @somearray : (@y = ("another","array"));'
$foo eq 'blah' ? (@x = @somearray) : (@y = ('another', 'array'));
-e syntax OK

Note the parentheses: ?: binds tighter than =.

Answer 2

The perlop documentation clearly states you should put parentheses around assignment operators.

Failing to use parentheses is a rod for your own back if you don't understand operator precedence. Stop trying to be too smart for your own good!

Answer 3

The Perl conditional operator is meant to be

$variable = (expression) ? true assignment : false assignment;

What you're doing looks like it should work and is basically the same as the if/else statement. But is just different enough from the norm to have issues.

Answer 4

This would be a good spot to use the lower precedence 'and' and 'or' operators.

$foo eq 'blah' and @x = @somearray or @y = ('another', 'array');

provided you are sure that @x = @somearray will always be true. or you could flip them around.

Answer 5

This is somewhat orthogonal to your question, but it bears pointing out: Perl's conditional operator propagates context from the first argument down into the second or third arguments, so this would give you undesired results:

$x = ($foo eq "blah") ? $somevalue : ("another","array");

If the conditional were false, $x would instead be assigned a single integer value 2 (the number of elements in the third argument).

If on the other hand you were attempting to perform a purely scalar assignment:

# this is wrong, for the same order-of-operations reasons as with arrays
($foo eq "blah") ? $x = $somevalue : $x = "another value";

This would be a reasonable (and the best) way to resolve the situation:

$x = ($foo eq "blah") ? $somevalue : "another value";

Likewise, you could optimize your original code this way:

@x = ($foo eq "blah") ? @somearray : ("another","array");