Arrays and Iteration

Array Sum (Linear Scan)

Walk an array once, accumulating each element into a running total. This is the canonical single-pass linear scan and the simplest possible loop invariant: after step i, total equals the sum of arr[0..i].

Algorithm

The canonical input from the lesson spec is arr=(3 1 4 1 5 9 2 6). After eight passes the running total is 31.

linear scan Visit each element exactly once in index order.
running total `total` accumulates the sum as the loop advances.

Basic Implementation

basic.sh
Replay: real traced execution (multi-file project)
#!/usr/bin/env bash
set -euo pipefail
arr=(3 1 4 1 5 9 2 6)
total=0
i=0
while [ "$i" -lt "${#arr[@]}" ]; do
	total=$((total + arr[i]))
	i=$((i + 1))
done
echo "$total"

  1. arr ← [3, 1, 4, 1, 5, 9, 2, 6]

    2set -euo pipefail3arr=(3 1 4 1 5 9 2 6)4total=0
    values this step[3, 1, 4, 1, 5, 9, 2, 6]arr

  2. total ← 0

    3arr=(3 1 4 1 5 9 2 6)4total=05i=0
    values this step0total[3, 1, 4, 1, 5, 9, 2, 6]arr

  3. i ← 0

    4total=05i=06while [ "$i" -lt "${#arr[@]}" ]; do
    values this step0i0total

  4. total ← 3

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step0 → 3total0i3arr[i]

  5. total ← 4

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step3 → 4total1i1arr[i]

  6. total ← 8

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step4 → 8total2i4arr[i]

  7. total ← 9

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step8 → 9total3i1arr[i]

  8. total ← 14

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step9 → 14total4i5arr[i]

  9. total ← 23

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step14 → 23total5i9arr[i]

  10. total ← 25

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step23 → 25total6i2arr[i]

  11. total ← 31

    6while [ "$i" -lt "${#arr[@]}" ]; do7	total=$((total + arr[i]))8	i=$((i + 1))
    values this step25 → 31total7i6arr[i]

Trace Output

trace.sh
Replay: real traced execution (multi-file project)
#!/usr/bin/env bash
set -euo pipefail
arr=(3 1 4 1 5 9 2 6)
total=0
i=0
while [ "$i" -lt "${#arr[@]}" ]; do
	before=$total
	total=$((total + arr[i]))
	printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
	i=$((i + 1))
done
printf 'final total = %d\n' "$total"

  1. total ← 3, stdout ← step 0: arr(0)=3 total 0 -> 3

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step3totalstep 0: arr(0)=3 total 0 -> 3stdout0before3arr[i]

  2. total ← 4, stdout ← step 1: arr(1)=1 total 3 -> 4

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step4totalstep 1: arr(1)=1 total 3 -> 4stdout3before1arr[i]

  3. total ← 8, stdout ← step 2: arr(2)=4 total 4 -> 8

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step8totalstep 2: arr(2)=4 total 4 -> 8stdout4before4arr[i]

  4. total ← 9, stdout ← step 3: arr(3)=1 total 8 -> 9

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step9totalstep 3: arr(3)=1 total 8 -> 9stdout8before1arr[i]

  5. total ← 14, stdout ← step 4: arr(4)=5 total 9 -> 14

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step14totalstep 4: arr(4)=5 total 9 -> 14stdout9before5arr[i]

  6. total ← 23, stdout ← step 5: arr(5)=9 total 14 -> 23

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step23totalstep 5: arr(5)=9 total 14 -> 23stdout14before9arr[i]

  7. total ← 25, stdout ← step 6: arr(6)=2 total 23 -> 25

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step25totalstep 6: arr(6)=2 total 23 -> 25stdout23before2arr[i]

  8. total ← 31, stdout ← step 7: arr(7)=6 total 25 -> 31

    7before=$total8total=$((total + arr[i]))9printf 'step %d: arr(%d)=%d total %d -> %d\n' "$i" "$i" "${arr[i]}" "$before" "$total"
    values this step31totalstep 7: arr(7)=6 total 25 -> 31stdout25before6arr[i]

  9. stdout ← final total = 31

    10	i=$((i + 1))11done12printf 'final total = %d\n' "$total"
    values this stepfinal total = 31stdout31total

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Bash: use the explicit while [ "$i" -lt "${#arr[@]}" ] loop with total=0 and a manual 0-based index. Bash has no built-in sum primitive; piping to awk '{ s += $1 } END { print s }' would push the loop into another process the lesson is not teaching, and printf '%s+' "${arr[@]}" 0 | bc would defer the running update to bc and hide it.
  • arr=(3 1 4 1 5 9 2 6) documents the fixed-content array; the manual i=$((i + 1)) step keeps the iteration without leaning on for v in "${arr[@]}" that hides the running index.
  • set -euo pipefail at the top keeps every typo and unset variable loud, and arithmetic uses $((...)) expansion so the assignments never trip set -e.
  • The replay shows i, arr[i], and total before and after each addition, matching the lesson spec's state-transition table.