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.rs
Replay: real traced execution (multi-file project)
fn main() {
	let arr = [3, 1, 4, 1, 5, 9, 2, 6];
	let mut total = 0;
	for i in 0..arr.len() {
		total = total + arr[i];
	}
	println!("{}", total);
}

  1. arr ← [3, 1, 4, 1, 5, 9, 2, 6]

    1fn main() {2	let arr = [3, 1, 4, 1, 5, 9, 2, 6];3	let mut total = 0;
    values this step[3, 1, 4, 1, 5, 9, 2, 6]arr

  2. total ← 0

    2let arr = [3, 1, 4, 1, 5, 9, 2, 6];3let mut total = 0;4for i in 0..arr.len() {
    values this step0total[3, 1, 4, 1, 5, 9, 2, 6]arr

  3. total ← 3

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step0 → 3total0i3arr[i]

  4. total ← 4

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step3 → 4total1i1arr[i]

  5. total ← 8

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step4 → 8total2i4arr[i]

  6. total ← 9

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step8 → 9total3i1arr[i]

  7. total ← 14

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step9 → 14total4i5arr[i]

  8. total ← 23

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step14 → 23total5i9arr[i]

  9. total ← 25

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step23 → 25total6i2arr[i]

  10. total ← 31

    4for i in 0..arr.len() {5	total = total + arr[i];6}
    values this step25 → 31total7i6arr[i]

Trace Output

trace.rs
Replay: real traced execution (multi-file project)
fn main() {
	let arr = [3, 1, 4, 1, 5, 9, 2, 6];
	let mut total = 0;
	for i in 0..arr.len() {
		let before = total;
		total = total + arr[i];
		println!("step {}: arr[{}]={} total {} -> {}",
			i, i, arr[i], before, total);
	}
	println!("final total = {}", total);
}

  1. total ← 3, stdout ← step 0: arr[0]=3 total 0 -> 3

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step3totalstep 0: arr[0]=3 total 0 -> 3stdout0before3arr[i]

  2. total ← 4, stdout ← step 1: arr[1]=1 total 3 -> 4

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step4totalstep 1: arr[1]=1 total 3 -> 4stdout3before1arr[i]

  3. total ← 8, stdout ← step 2: arr[2]=4 total 4 -> 8

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step8totalstep 2: arr[2]=4 total 4 -> 8stdout4before4arr[i]

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

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step9totalstep 3: arr[3]=1 total 8 -> 9stdout8before1arr[i]

  5. total ← 14, stdout ← step 4: arr[4]=5 total 9 -> 14

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step14totalstep 4: arr[4]=5 total 9 -> 14stdout9before5arr[i]

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

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step23totalstep 5: arr[5]=9 total 14 -> 23stdout14before9arr[i]

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

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step25totalstep 6: arr[6]=2 total 23 -> 25stdout23before2arr[i]

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

    5let before = total;6total = total + arr[i];7println!("step {}: arr[{}]={} total {} -> {}",
    values this step31totalstep 7: arr[7]=6 total 25 -> 31stdout25before6arr[i]

  9. stdout ← final total = 31

    9	}10	println!("final total = {}", total);11}
    values this stepfinal total = 31stdout31total

Complexity

  • Time: O(n)
  • Space: O(1)

Implementation notes

  • Rust: use the explicit for i in 0..arr.len() loop with let mut total = 0;. The iterator form arr.iter().sum() is fine for production but hides the loop the lesson spec is teaching.
  • let arr = [3, 1, 4, 1, 5, 9, 2, 6]; documents the fixed-size [i32; 8] array contract; arr.len() gives the explicit count without leaning on a helper that hides the iteration.
  • The replay shows i, arr[i], and total before and after each addition, matching the lesson spec's state-transition table.