Array traversal in C++ - Step-by-Step Execution

Concept Flow - Array traversal

Start at index 0

↓

Check: index < array_size?

No→EXIT

Yes↓

Access array[index

↓

Process element

↓

Increment index

↩Back to Check

We start at the first element (index 0), check if the index is within the array size, process the element, then move to the next index until we reach the end.

Execution Sample

C++

#include <iostream>
using namespace std;

int main() {
    int arr[] = {10, 20, 30};
    int n = 3;
    for (int i = 0; i < n; i++) {
        cout << arr[i] << " ";
    }
    return 0;
}

This code prints each element of the array one by one.

Execution Table

Iteration	i	Condition (i < n)	Action	Output
1	0	True	Print arr[0] = 10	10
2	1	True	Print arr[1] = 20	20
3	2	True	Print arr[2] = 30	30
4	3	False	Exit loop

💡 i reaches 3, condition 3 < 3 is False, loop ends

Variable Tracker

Variable	Start	After 1	After 2	After 3	Final
i	undefined	0	1	2	3
Output	""	"10 "	"10 20 "	"10 20 30 "	"10 20 30 "

Key Moments - 3 Insights

Why does the loop stop when i equals the array size?

Because the condition i < n becomes false at i = 3 (array size), so the loop exits as shown in execution_table row 4.

Why do we start indexing from 0 instead of 1?

Arrays in C++ start at index 0, so the first element is arr[0]. This is why the loop starts with i = 0 (execution_table row 1).

What happens if we try to access arr[i] when i >= n?

Accessing arr[i] when i >= n causes undefined behavior (out of bounds). The loop condition prevents this by stopping before i reaches n (see exit condition in execution_table).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of i during the third iteration?

A 1

B 2

C 3

D 0

💡 Hint

Check the 'i' column in the third row of the execution_table.

At which iteration does the loop condition become false?

A Iteration 4

B Iteration 3

C Iteration 2

D Iteration 1

💡 Hint

Look at the 'Condition (i < n)' column where it becomes False in the execution_table.

If the array size n was 4 instead of 3, how many times would the loop run?

A 5 times

B 3 times

C 4 times

D 2 times

💡 Hint

The loop runs while i < n, so if n=4, it runs for i=0,1,2,3 (4 times).

Concept Snapshot

Array traversal in C++:
Use a for loop from i = 0 to i < array_size.
Access elements with arr[i].
Loop stops before i reaches array_size to avoid out-of-bounds.
Print or process each element inside the loop.

Full Transcript

This example shows how to traverse an array in C++. We start with index i at 0, check if i is less than the array size n, then access and print the element at arr[i]. After processing, we increase i by 1 and repeat. The loop stops when i equals n, preventing out-of-bounds access. The output prints each element separated by spaces.