Loop patterns, sums and star patterns · C++

A loop on its own just repeats. Most real loops build up an answer one step at a time: a total, a count, a product. These are the patterns you'll meet in class and in exams; run each one and change the numbers.

Iteration and the sum of 1 to 10

Each pass through a loop's body is an iteration. To add numbers up, keep a running total in an accumulator: start it at 0 before the loop, add to it inside, and use it after:

C++

#include <iostream>
using namespace std;

int main() {
    int sum = 0;                         // 1. start the total at 0
    for (int i = 1; i <= 10; i++) {
        sum += i;                        // 2. add this number to the total
        cout << "iteration " << i << ": total is " << sum << endl;
    }
    cout << "Sum of 1 to 10 = " << sum << endl;   // 3. use it after the loop
    return 0;
}

i	sum before	sum after
1	0	1
2	1	3
3	3	6
...	...	...
10	45	55

When i becomes 11, i <= 10 is false and the loop stops. Forgetting = 0 on sum leaves it holding a random leftover value.

Counting down: 10 to 1 in reverse

Start high, continue while i >= 1, and subtract with i--. Adding while counting down gives the same 55:

C++

#include <iostream>
using namespace std;

int main() {
    int sum = 0;
    for (int i = 10; i >= 1; i--) {
        cout << i << " ";
        sum += i;
    }
    cout << endl << "Sum of 10 down to 1 = " << sum << endl;

    int n = 5;
    while (n >= 1) {               // the same countdown with while
        cout << n << "... ";
        n--;
    }
    cout << "lift off!" << endl;
    return 0;
}

Conditional sums, counting and factorials

C++

#include <iostream>
using namespace std;

int main() {
    int evens = 0, odds = 0, sevens = 0;
    for (int i = 1; i <= 10; i++) {
        if (i % 2 == 0) evens += i;
        else odds += i;
    }
    for (int i = 1; i <= 100; i++) {
        if (i % 7 == 0) sevens++;          // a counter adds 1 each time
    }
    int fact = 1;                          // a product starts at 1, not 0
    for (int i = 1; i <= 10; i++) fact *= i;
    cout << "Even sum: " << evens << ", odd sum: " << odds << endl;
    cout << sevens << " numbers from 1 to 100 divide by 7" << endl;
    cout << "10! = " << fact << endl;
    return 0;
}

Times tables, array totals and digits

C++

#include <iostream>
using namespace std;

int main() {
    int n = 7;
    for (int i = 1; i <= 10; i++) {
        cout << n << " x " << i << " = " << n * i << endl;
    }

    int marks[5] = {72, 85, 90, 64, 88};
    int total = 0, highest = marks[0];
    for (int i = 0; i < 5; i++) {
        total += marks[i];
        if (marks[i] > highest) highest = marks[i];
    }
    cout << "Total: " << total << ", highest: " << highest << endl;

    int num = 4096, digitSum = 0, reversed = 0;
    while (num > 0) {
        int digit = num % 10;              // the last digit
        digitSum += digit;
        reversed = reversed * 10 + digit;
        num /= 10;                         // drop the last digit
    }
    cout << "Digit sum: " << digitSum << ", reversed: " << reversed << endl;
    return 0;
}

Patterns with nested loops

The outer loop picks the row; the inner loop prints that row. A triangle row has row stars; a pyramid row has n - row spaces then 2 * row - 1 stars; Floyd's triangle keeps a counter outside both loops:

C++

#include <iostream>
using namespace std;

int main() {
    int n = 5;
    for (int row = 1; row <= n; row++) {
        for (int col = 1; col <= row; col++) cout << "* ";
        cout << endl;
    }
    for (int row = 1; row <= n; row++) {
        for (int s = 1; s <= n - row; s++) cout << " ";
        for (int k = 1; k <= 2 * row - 1; k++) cout << "*";
        cout << endl;
    }
    int num = 1;
    for (int row = 1; row <= 4; row++) {
        for (int col = 1; col <= row; col++) {
            cout << num << " ";
            num++;
        }
        cout << endl;
    }
    return 0;
}

Your turn: sum of 1 to n

This example reads n with cin; its input is 100:

C++

#include <iostream>
using namespace std;

int main() {
    int n, sum = 0;
    cout << "Add up 1 to which number? ";
    cin >> n;
    for (int i = 1; i <= n; i++) sum += i;
    cout << endl << "1 + 2 + ... + " << n << " = " << sum << endl;
    cout << "Formula check: " << n * (n + 1) / 2 << endl;
    return 0;
}