AP CSA Unit 1 Day 7: Object State Over Time Via Methods
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Advanced Practice Question
String s = "COMPUTER";
String t = s.substring(0, 4);
String u = t + s.substring(6);
int x = u.indexOf("ER");
System.out.println(u + " " + x);Trace each line using the String methods from the Java Quick Reference:
Line 1: s = "COMPUTER"
Indices: C=0, O=1, M=2, P=3, U=4, T=5, E=6, R=7
Line 2: t = s.substring(0, 4) → characters at indices 0, 1, 2, 3 → "COMP"
Remember: substring(start, end) includes start but excludes end.
Line 3: s.substring(6) → characters from index 6 to the end → "ER"
→ u = "COMP" + "ER" = "COMPER"
Line 4: u.indexOf("ER") → searches "COMPER" for "ER"
Indices of u: C=0, O=1, M=2, P=3, E=4, R=5
→ "ER" starts at index 4 in "COMPER"
Output: COMPER 4
B) COMPUTER 6 — This assumes substring(0, 4) includes index 4 (the 'U'). It does not — the end index is exclusive. Also, indexOf would search the wrong string.
C) COMPER 6 — Gets the string correct but searches for "ER" in the original "COMPUTER" (index 6) instead of in "COMPER" (index 4). The indexOf is called on u, not s.
D) COMP 4 — Forgets the concatenation on line 3 and incorrectly assumes u is just "COMP".
Two critical traps here: (1) substring(0, 4) returns 4 characters at indices 0-3, not 0-4. The end index is always exclusive. (2) indexOf searches the String it is called on (u), not the original String (s). Students who get "COMPER" correct but answer 6 are searching the wrong variable.
When tracing String problems, write out the index numbers above each character. For substring(a, b), circle characters from index a up to (but not including) index b. For indexOf, always verify which String you are searching — the AP exam loves to test whether you track which variable holds which value.
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