← IGCSE Computer Science Paper 2

Looping with a lower bound of zero

For ARRAY[0:29], the loop must start at the lower bound 0 and end at the upper bound 29. It then visits all 30 elements and never goes outside the array.

Part of: IGCSE Computer Science 0478 — 8.2 Arrays — Module 17

Understanding Looping with a lower bound of zero in IGCSE Computer Science Paper 2

What it does: The array is declared with bounds 0 and 9, so the loop limits must be 0 and 9 to match the declaration. ARRAY[0:49] has a lower bound of 0 and an upper bound of 49, so the FOR loop uses exactly those two numbers. Indexes 0 to 29 inclusive give 30 elements (29 - 0 + 1 = 30). The loop visits each one once and never leaves the array.

Common mistakes: Mark has a lower bound of 0, so the loop must start at 0. Starting at 1 skips Mark[0]. The correct line is FOR I ← 0 TO 9. The array runs from index 0 to 29. A loop starting at 1 misses element 0. The correct line is FOR Index ← 0 TO 29.

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This module's key terms are also in a printable crossword (Modules 459–468) and word search (Modules 459–478).

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Learning

Duo

An array is declared as ARRAY[0:29] OF STRING. Which loop visits every element exactly once?

FOR Index ← 0 TO 29FOR Index ← 1 TO 30

Matches

FOR Index ← 1 TO 29 (for ARRAY[0:29])

↓

The loop starts at 1, so element 0 is never visited. It should start at the lower bound 0.

Applying

Meanings

Term

FOR Index ← 0 TO 29

Meaning

For ARRAY[0:29], the loop must start at the lower bound 0 and end at the upper bound 29. It then visits all 30 elements and never goes outside the array.

Practice

DECLARE Scores : ARRAY[0:29] OF INTEGER

Testing

Gaps

FOR Index ← TO 29

Check Answer

Spots

1// Mark is ARRAY[0:9]

2FOR I ← 1 TO 9

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Quick Questions

1. An array is declared as ARRAY[0:29] OF STRING. Which loop visits every element exactly once?

A. FOR Index ← 1 TO 30

B. FOR Index ← 0 TO 29 ✓

The loop must start at the lower bound 0 and finish at the upper bound 29. Starting at 1 misses element 0 and ending at 30 goes past the end.

2. DECLARE Scores : ARRAY[0:9] OF INTEGER. Which loop header matches this declaration?

A. FOR Index ← 0 TO 9 ✓

B. FOR Index ← 1 TO 10

The loop limits must match the declaration: lower bound 0 and upper bound 9. Index 10 does not exist, so the second loop would cause an error.

3. A student uses FOR Index ← 1 TO 30 on ARRAY[0:29]. What goes wrong?

A. It skips element 0 and tries to use element 30, which is outside the array ✓

B. It works correctly because there are 30 loop repeats

The loop does run 30 times, but the indexes are 1 to 30. Index 0 is never visited and index 30 is out of range. Counting repeats is not enough; the index values must match the bounds.

4. For ARRAY[0:29], which loop starts at the lower bound?

A. FOR Index ← 0 TO 29 ✓

B. FOR Index ← 29 TO 0

A FOR loop counts up from the first value to the last. The lower bound 0 goes first and the upper bound 29 goes after TO.

5. Names is declared as ARRAY[0:4] OF STRING. Which loop goes outside the array?

A. FOR Index ← 0 TO 4

B. FOR Index ← 0 TO 5 ✓

The upper bound is 4, so the loop must stop at 4. Going to 5 would access Names[5], which does not exist.

Match the Pairs

FOR Index ← 0 TO 29 → Visits every element of ARRAY[0:29] from the lower bound to the upper bound

FOR Index ← 1 TO 30 → Misses element 0 and goes past the end of ARRAY[0:29]

To loop through ARRAY[0:9], start at the lower bound and finish at the upper bound → FOR Index ← 0 TO 9

The loop limits must match the declaration of the array ARRAY[0:49] → FOR Index ← 0 TO 49

FOR Index ← 1 TO 29 (for ARRAY[0:29]) → The loop starts at 1, so element 0 is never visited. It should start at the lower bound 0.

FOR Index ← 0 TO 30 (for ARRAY[0:29]) → The loop ends at 30, which is past the upper bound 29, so it goes out of range. It should end at 29.

Key Term

FOR Index ← 0 TO 29

For ARRAY[0:29], the loop must start at the lower bound 0 and end at the upper bound 29. It then visits all 30 elements and never goes outside the array.

Fill in the Blanks

1. DECLARE Scores : ARRAY[0:29] OF INTEGER FOR Index ← 0 TO 29

Hint: Type the starting value of the loop, which must match the lower bound of the array.

2. DECLARE Names : ARRAY[0:29] OF STRING FOR Index ← 0 TO 29

Hint: Type the final value of the loop, which must match the upper bound in the declaration.

3. DECLARE Marks : ARRAY[0:9] OF INTEGER FOR Index ← 0 TO 9 OUTPUT Marks[Index]

Hint: Type the loop limit that makes the loop stop at the last element of this array.

4. DECLARE Items : ARRAY[0:19] OF STRING FOR Index ← 0 TO 19

Hint: Type the lower bound that the loop start value is matching.

5. FOR Index ← 0 TO 29 OUTPUT Scores[Index] NEXT Index

Hint: Type the keyword that separates the start value from the end value in a FOR loop.

Spot the Error

Question 1

A // Mark is ARRAY[0:9]

B FOR I ← 1 TO 9

C Mark[I] ← 0

D NEXT I

Line B — Mark has a lower bound of 0, so the loop must start at 0. Starting at 1 skips Mark[0]. The correct line is FOR I ← 0 TO 9.

Question 2

A // Mark is ARRAY[0:29]

B FOR Index ← 1 TO 29

C OUTPUT Mark[Index]

D NEXT Index

Line B — The array runs from index 0 to 29. A loop starting at 1 misses element 0. The correct line is FOR Index ← 0 TO 29.

Question 3

A // Temp is ARRAY[0:29]

B Total ← 0

C FOR Index ← 0 TO 30

D Total ← Total + Temp[Index]

E NEXT Index

Line C — The upper bound is 29. Going up to 30 reads Temp[30], which is outside the array. The correct line is FOR Index ← 0 TO 29.

Question 4

A // Score is ARRAY[0:29]

B FOR Index ← 0 TO 28

C Score[Index] ← 0

D NEXT Index

Line B — The loop stops one element early, so Score[29] is never set. It must run to the upper bound: FOR Index ← 0 TO 29.

Question 5

A DECLARE T : ARRAY[0:8] OF REAL

B FOR I ← 0 TO 9

C T[I] ← 0.5

D NEXT I

Line A — The loop runs from 0 to 9, so the array needs an upper bound of 9. With [0:8], T[9] would be out of range. The declaration should be ARRAY[0:9].

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