To answer this question, let's break down the code and go through each option:
The code initializes a scalar variable $val with a value of 'x'.
The ref() function in Perl is used to determine the reference type of a variable. If a variable is not a reference, ref() will return an empty value.
Option A) SCALAR - This option is incorrect because ref() does not return the reference type of the variable in this case. It will return an empty value.
Option B) empty value - This option is correct because ref() will return an empty value since $val is not a reference.
Option C) STRING - This option is incorrect because ref() does not return the reference type of the variable in this case. It will return an empty value.
Option D) "not a reference" - This option is incorrect because the code does not explicitly output the string "not a reference". It will return an empty value.
The correct answer is B. The code will print an empty value.