Given: ``java class Test { int al; public static void doProduct(int a) { a = a * a; } public static void doString(StringBuilder s) { s.append(" " + s); } public static void main(String[] args) {…

Question

class Test {
 int al;
 public static void doProduct(int a) {
 a = a * a;
 }
 public static void doString(StringBuilder s) {
 s.append(" " + s);
 }
 public static void main(String[] args) {
 Test item = new Test();
 item.al = 11;
 StringBuilder sb = new StringBuilder("Hello");
 Integer i = 10;
 doProduct(i);
 doString(sb);
 doProduct(item.al);
 System.out.println(i + " " + sb + " " + item.al);
 }
}

Options

• A10 Hello Hello 11
• B10 Hello Hello 121
• C100 Hello 121
• D10 Hello 11
• E100 Hello Hello 121

• A
  70% (21)
• B
  3% (1)
• C
  3% (1)
• D
  17% (5)
• E
  7% (2)

Explanation

Option A is correct because Java passes primitives by value and objects by reference - but "by reference" means the reference itself is copied, not the object's address being rebindable.

• doProduct(i): Integer i is unboxed to a primitive int when passed. The method mutates only its local copy a, so i stays 10.
• doString(sb): StringBuilder is mutable and passed by reference, so s.append(" " + s) actually modifies sb. At call time, s evaluates to "Hello", so " " + s = " Hello", making sb become "Hello Hello".
• doProduct(item.al): item.al (an int) is passed by value; the field stays 11.

Why each distractor fails:

• B (...121): Assumes item.al was modified - it wasn't; primitives are copied.
• C (100 Hello 121): Assumes i changed to 100 and sb wasn't modified - both wrong.
• D (10 Hello 11): Assumes StringBuilder wasn't modified - it was, because StringBuilder is a mutable object passed by reference.
• E (100 Hello Hello 121): Assumes both i and item.al changed - neither did.

Memory tip: Think "PRIM = private copy, Object = shared." Primitives (and unboxed wrappers like Integer) get their own copy in a method; mutable objects like StringBuilder share the underlying data, so changes inside a method stick.

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