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Which of the following are output by the following code? (Choose all that apply)public class StringBuilders {        public static StringBuilder work(StringBuilder a, StringBuilder b) {               a = new StringBuilder("a");               b.append("b");               return a;             }            public static void main(String[] args) {                  StringBuilder s1 = new StringBuilder("s1");                  StringBuilder s2 = new StringBuilder("s2");                  StringBuilder s3 = work(s1, s2);                  System.out.println("s1 = " + s1);                  System.out.println("s2 = " + s2);                  System.out.println("s3 = " + s3);                }      }*A. s1 = aB. s1 = s1C. s2 = s2D. s2 = s2bE. s3 = aF. s3 = nullG. The code does not compile.

Question

Which of the following are output by the following code? (Choose all that apply)public class StringBuilders {        public static StringBuilder work(StringBuilder a, StringBuilder b) {               a = new StringBuilder("a");               b.append("b");               return a;             }            public static void main(String[] args) {                  StringBuilder s1 = new StringBuilder("s1");                  StringBuilder s2 = new StringBuilder("s2");                  StringBuilder s3 = work(s1, s2);                  System.out.println("s1 = " + s1);                  System.out.println("s2 = " + s2);                  System.out.println("s3 = " + s3);                }      }*A. s1 = aB. s1 = s1C. s2 = s2D. s2 = s2bE. s3 = aF. s3 = nullG. The code does not compile.

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Solution

The output of the code will be:

B. s1 = s1 D. s2 = s2b E. s3 = a

Here's why:

In the work method, a new StringBuilder "a" is assigned to a, but this does not change the original s1 because Java is pass-by-value, meaning it gets a copy of the reference to the StringBuilder, not the reference itself. So, s1 remains "s1".

The append method is used on b which affects the original s2 because methods called on the object will modify the original object. So, s2 becomes "s2b".

The work method returns the new StringBuilder "a", which is assigned to s3. So, s3 is "a".

This problem has been solved

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