Given the code fragment: String s = "Hello Java"; System.out.println (s.length()); s.concat ("SE 8"); System.out.println (s.length ()); What is the result?

The correct answer is B. 10 10. Option B is correct because Java String objects are immutable - s.concat("SE 8") returns a new String object but does not modify s, and since the return value is discarded (not assigned back to s), s still points to "Hello Java" (length 10) for both println calls. Why…

Data Types and Operators

Question

Given the code fragment:
String s = "Hello Java";
System.out.println (s.length());
s.concat ("SE 8");
System.out.println (s.length ());
What is the result?

Options

A12 10
B10 10
C9 9
D10 14

How the community answered

(50 responses)

A
2% (1)
B
92% (46)
C
2% (1)
D
4% (2)

Explanation

Option B is correct because Java String objects are immutable - s.concat("SE 8") returns a new String object but does not modify s, and since the return value is discarded (not assigned back to s), s still points to "Hello Java" (length 10) for both println calls.

Why distractors are wrong:

A (12, 10): 12 is not the length of any string involved here; "Hello Java" has exactly 10 characters.
C (9, 9): Off-by-one error - "Hello Java" is 10 characters including the space, not 9.
D (10, 14): This would be correct if the code read s = s.concat("SE 8") - "Hello JavaSE 8" is 14 characters - but without reassignment, s is never updated.

Memory tip: In Java, think of String methods as returning a receipt - if you don't save the receipt (s = s.concat(...)), the transaction is lost. The original string never changes.

Topics

#String immutability#String concatenation#String.length()#String methods

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