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8. Week 4 Thursday: Loop Management

Last week, we talked about writing for loops and while loops, and on Tuesday, we saw some examples on how to apply these skills to reformatting strings.

There’s a slight gap in our looping skills. A lot of turn-based video games rely heavily on loops. They might be set up as, “While the player has more than 0 health….”, “While the player still has cards in the deck…”, etc. However, one may imagine that players can forfeit the game, pass their turn, or play cards that modify the flow of the game.

A specific example may be Texas Hold’em, which has some fairly complicated logic in how each round is structured. Players that fold their hand early on shouldn’t be asked to bet in the same round, but they can still play in the next. In other words, some players are skipped over while others play, and handling this with a loop can become quite delicate.

Let’s consider the following simpler problem to illustrate what we mean:

Sample Run: 
INPUT   1 3 5 7 9
OUTPUT  25

INPUT   1 2 
OUTPUT  1 

We may write some pseudocode for this as follows:

1. Initialise a running sum to 0.
2. Repeat the following five times:
   • Take a number from the user as input.
   • If the user inputs an odd number, add it to the running sum.
   • If the user inputs an even number, skip the rest of the repetitions of this step.
3. Print out the running sum.

In C++, this becomes:

 1#include <iostream>     // needed for cin, cout, and endl
 2
 3using namespace std; 
 4
 5int main() {
 6    // 1. initialise the running sum. 
 7    int running_sum = 0; 
 8
 9    // 2. 5 repetitions of...
10    for(int i = 0; i < 5; i++) {
11        // 2a. get user input 
12        int n = 0; 
13        cin >> n; 
14
15        // 2b. if input is odd, add to sum 
16        if(n % 2 == 1) {
17            running_sum += n; 
18        }
19
20        // 2c. otherwise, exit the loop 
21        else {
22            // ????? 
23        }
24    }
25
26    // 3. print results 
27    cout << running_sum << endl; 
28    return 0; 
29}

How do we fill in line 22? It’s certainly not impossible to modify the above program using only what we’ve learned so far. For instance, one could create a boolean variable that keeps track of whether or not the user has inputted an even number yet. Then, one could wrap everything within the for loop inside an if statement. This is bulky, tedious, and overcomplicated, and moreover, it adds a lot of extra overhead for us programmers. There’s a better way to do this:

The break and continue statements

There are two ways to alter loops from within a loop:

• continue
  Skips the remaining code within a loop and starts the next repetition.
• break
  Exits a loop prematurely and skips all future repetitions.

The break statement functions the same way in both for and while loops: it just “jumps” to the end of the loop, no questions asked. However, the continue statement has a very subtle difference between the two loops. Consider the following four examples, where a loop is set up to print the numbers 0 through 9. Once it hits the number 5, however, the code either runs a break statement or a continue statement.

 1/* 1. break in a for loop */
 2for(int i = 0; i < 10; i++) {
 3    if(i == 5) {
 4        break; 
 5    }
 6
 7    cout << i << endl; 
 8}
 9
10/* 2. break in a while loop */ 
11int i = 0; 
12while(i < 10) {
13    if(i == 5) {
14        break; 
15    }
16
17    cout << i << endl; 
18    i++;    // update the counter! 
19} 
20
21/* 3. continue in a for loop */
22for(int i = 0; i < 10; i++) {
23    if(i == 5) {
24        continue; 
25    }
26
27    cout << i << endl; 
28}
29
30/* 4. continue in a while loop */ 
31int i = 0; 
32while(i < 10) {
33    if(i == 5) {
34        continue; 
35    }
36
37    cout << i << endl; 
38    i++;    // update the counter! 
39}

Both the first and the second examples will print the numbers 0 through 4, each on a separate line. Once it reaches the sixth iteration of the loop (i.e., when i == 5), the break statement is run, which causes the loop to end immediately.

The third example will print all of the numbers between 0 and 9 except for 5, each on a separate line. When i == 5, the continue statement is run. Instead of skipping to the end of the entire loop, it skips to the end of the current iteration. The update condition i++ is run, and the rest of the loop runs normally.

The fourth example will print all of the numbers between 0 and 4, each on a separate line, then enter an infinite loop. This is because when the program reads the continue statement, it skips to the end of the current iteration. This also skips the update step, i++, and hence i is forever stuck with a value of 5.

Let’s now return to the example we started with. We may fill in the missing line of our code as follows:

 1#include <iostream>     // needed for cin, cout, and endl
 2
 3using namespace std; 
 4
 5int main() {
 6    // 1. initialise the running sum. 
 7    int running_sum = 0; 
 8
 9    // 2. 5 repetitions of...
10    for(int i = 0; i < 5; i++) {
11        // 2a. get user input 
12        int n = 0; 
13        cin >> n; 
14
15        // 2b. if input is odd, add to sum 
16        if(n % 2 == 1) {
17            running_sum += n; 
18        }
19
20        // 2c. otherwise, exit the loop 
21        else {
22            break; 
23        }
24    }
25
26    // 3. print results 
27    cout << running_sum << endl; 
28    return 0; 
29}

Notice the introduction of the break statement on line 22. We can even rearrange the code as follows, which is more concise but deviates slightly from our pseudocode:

 1#include <iostream>     // needed for cin, cout, and endl
 2
 3using namespace std; 
 4
 5int main() {
 6    // 1. initialise the running sum. 
 7    int running_sum = 0; 
 8
 9    // 2. 5 repetitions of...
10    for(int i = 0; i < 5; i++) {
11        // 2a. get user input 
12        int n = 0; 
13        cin >> n; 
14
15        // exit the loop immediately for even numbers 
16        if(n % 2 == 0) {
17            break; 
18        }
19
20        // add n to the running sum 
21        running_sum += n; 
22    }
23
24    // 3. print results 
25    cout << running_sum << endl; 
26    return 0; 
27}

If n is odd, the break statement on line 17 will not be run, and it will be added to the running sum. If n is even, it will run the break statement and skip over line 21, as well as the rest of the loop!

Practise Problems — Predict the Output

As with everything in this class, getting better at a skill involves practise. Many relatively simple problems don’t necessitate the use of break and continue, and those that do can be circumvented with well-constructed if/else-if/else statements. So, rather than spending hours trying to engineer a situation like that, we’ll get some practise understanding how these statements function and keep these tools in our back pocket for when they organically arise in the future.

For the following three problems, predict the outcome of the code. Try not to copy and paste these into your IDE and run them on your computer unless you get stuck.

 1#include <iostream>
 2#include <string> 
 3
 4using namespace std; 
 5
 6int main() {
 7    string s = "hbDe8391cd31de"; 
 8
 9    for(int i = 0; i < s.length(); i++) {
10        char c = s.at(i); 
11        if('a' <= c && c <= 'z') {
12            continue; 
13        }
14
15        cout << c << endl; 
16        if('A' <= c && c <= 'Z') {
17            break; 
18        }
19    }
20
21    return 0; 
22}

 1#include <iostream> 
 2
 3using namespace std; 
 4
 5int main() {
 6    int n = 245; 
 7
 8    for(int i = 1; i <= n; i++) {
 9        if(n % i == 0) {
10            cout << n / i << endl; 
11
12            if(n % (i * i) == 0) {
13                break; 
14            } else {
15                continue; 
16            }
17        }
18
19        cout << i << endl; 
20    }
21
22    return 0; 
23}

 1#include <iostream> 
 2
 3using namespace std; 
 4
 5int main() {
 6    int a = 637; 
 7    int b = 780; 
 8
 9    int n = 780; 
10    while(n > 0) {
11        if(b % n != 0) {
12            n--; 
13            continue; 
14        }
15
16        if(a % n == 0) {
17            break; 
18        }
19
20        n--; 
21    }
22
23    cout << n << endl; 
24    return 0; 
25}