Given the code fragment: 7. StringBuilder sb1 = new StringBuilder("Duke"); 8. String str1 = sb1.toString(); 9. // insert code here 10. System.out.print(str1 == str2); Which code fragment, when…

Question

Options

• AString str2 = sb1.toString();
• BString str2 = str1;
• CString str2 = "Duke";
• DString str2 = new String(str1);

• A
  12% (3)
• B
  76% (19)
• C
  8% (2)
• D
  4% (1)

Explanation

Option B works because str2 = str1 makes both variables point to the same object in memory - == on objects checks reference identity (same address), not value equality, so two references to the same object always return true.

Why the distractors fail:

• A (sb1.toString() again) - toString() creates a new String object each call, so str1 and str2 point to different heap objects even though their content is identical.
• C ("Duke" literal) - String literals live in the string pool, but str1 was created via toString() which allocates a new heap object outside the pool; the two objects are not the same reference.
• D (new String(str1)) - The new keyword explicitly creates a fresh heap object, so it can never be == to str1 even though it copies the content.

Memory tip: Think of == as asking "are these the same physical box?" and .equals() as asking "do these boxes contain the same thing?" - only assignment (str2 = str1) guarantees both variables point to the exact same box.

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