Loops

03603111 Programming Fundamentals I

Department of Computer Engineering, Faculty of Engineering at Sriracha

Course outline and schedule

• L01 Introduction
• L02 Data and I/O
• L03 Conditionals
• L04 Loops (today!)
• L05 Arrays, strings, and pointers
• L06 Functions and abstraction
• Lab exam #1

• L07 Nested loops, file I/O, and dynamic arrays
• L08 Sorting
• L09 Searching
• L10 Structures
• L11 Linked data structures
• L12 Unions and tagged structures
• Lab exam #2

2

Overview

• The while loop
• Loop invariants
• Loop interruption: break and continue
• The do … while loop

3

Previous lesson recaps

4

if statement – do it or skip it

• What you’ve seen in the previous example is the if statement

​

Syntax: if (condition) statement

​

• The condition is a Boolean expression

5

if … else statement – do this or do that

Syntax:

if (condition) statement₁ else statement₂

​

int n = 123;

if (n % 2 == 0)

printf("%d is even.\n", n);

else

printf("%d is odd.\n", n);

​

~> 123 is odd.

6

Comparison and logical operators

• Boolean expressions compare values or expressions using comparison operators
• Comparison operators
• Comparison operators yield 1 for true and 0 for false
• Note the low precedence of these operators, compared to arithmetic operators

​

• Multiple conditions can be combined using logical operators
• Logical operators
• We use them to combine “logical clauses”
• && has higher precedence than ||, and ! has the highest precedence

7

Multiway selection – the if … else if … else pattern

#include <stdio.h>

​

int main()

{

char c;

​

printf("Enter a character: ");

scanf(" %c", &c); // note the space before %c

​

if (c >= 'A' && c <= 'Z') {

printf("'%c' is a capital letter.\n", c);

} else if (c >= 'a' && c <= 'z') {

printf("'%c' is a small letter.\n", c);

} else {

printf("'%c' is not a letter.\n");

}

}

8

Enter a character: x

'x' is a small letter.

​

// Re-run

Enter a character: A

'A' is a capital letter.

​

// Re-run

Enter a character: 9

'9' is not a letter.

Results

General cautions and guidelines

• Never use = (assignment) where == (equality) should be used

​

• Never compare floating-points for equality
• Check if the value is within tolerance (e.g., 0.0001f) rather than being exactly equal (fabs() is helpful here)

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• When there are two actions to choose from, use if … else

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• When there are two values to choose from, use conditional expressions (cond ? a : b)

9

The U-SWACT Problem Solving Framework

1. ͏UNDERSTAND the problem
• What is the task really asking?
• What are the inputs and expected outputs?
• Are there any constraints (e.g., size, time, types, value ranges)?
2. Make up SAMPLE INPUTS
• Try normal and edge cases
• Include small, negative, or boundary values
3. ͏WORK OUT solutions manually
• For each sample input, figure out the correct output by hand
• Explain your reasoning in plain language
• Look for patterns or repeated steps
• If stuck, simplify the problem or solve a related problem first

4. Write the ALGORITHM
• Write down the steps from input to output
• Break it into sub-problems if needed
• Sketch out any loops or conditions
5. Translate into CODE
• Start small – implement one step at a time
• Test frequently as you build
• Use print or a debugger to trace values
• If it doesn’t work, go back to your algorithm or manual steps
6. ͏TEST the code
• Try all your sample inputs from Step 2
• Add more: normal, edge, and invalid inputs
• Ask: “Does this cover all cases I can think of?”

10

The while loop

11

Let’s start with problems again

• Many problems require us to do things repeatedly
• Compute 10!
• Find the maximum among all the integers entered by the user
• Find the largest N such that the sum from 1 to N is still less than 50

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• Some are task-oriented
• Given N, print N stars
• Keep reading the user input until the input is within the range 1-100

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• Take a minute to think how a computer can solve these problems
• Again, remember that we cannot process all the numbers at once
• Let’s say we do it one at a time, but repeatedly

Conditional loops

• There are 3 loop (iteration) control structures in C
• while loop
• do … while loop
• for loop

​

• All of them are conditional – meaning that they will repeatedly do things until the condition is false

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• We’ll focus only on the while and do … while loops for now

13

while loop – check first, then act

• The first is while loop
• Syntax: while (condition) statement

​

• The while loop repeatedly executes statement until condition is false
• The condition is checked first – if it’s false from the beginning, statement will never be executed (no loop happening)

14

Condition

Statement(s)

False

True

Problem: Printing N stars

#include <stdio.h>

​

int main()

{

int n;

​

printf("How many stars do you want? ");

scanf("%d", &n);

​

while (n > 0) {

printf("*");

n -= 1;

}

printf("\n");

​

return 0;

}

15

How many stars do you want? 5

*****

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// Re-run

How many stars do you want? 0

Result:

Introducing the (in)famous ++ and -- operators

++ is the increment operator, and -- is the decrement operator

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• C++ got its name from this operator!
• Can’t be used with floating-point types

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• Both a++ and ++a by themselves are similar to a += 1
• When it’s used in an expression or assignment, the difference is important
• ++a increments a first, then uses the incremented value in the expression
• a++ uses the current value of a in the expression first, then increments a
• Given a = 10:
• b = a++ * 2; // a == 11, b == 20
• b = ++a * 2; // a == 11, b == 22

16

Problem: Printing N stars using -- operator

Direct substitution

int main()

{

int n;

​

printf("How many stars … ");

scanf("%d", &n);

​

while (n > 0) {

printf("*");

n--; // replaces n -= 1;

}

printf("\n");

}

Concise and idiomatic usage

int main()

{

int n;

​

printf("How many stars … ");

scanf("%d", &n);

​

// Check and decrease at once

while (n-- > 0) {

printf("*");

}

printf("\n");

}

17

Similar problem: Counting up to N

#include <stdio.h>

​

int main()

{

int n;

​

printf("Number to count up to: ");

scanf("%d", &n);

​

int v = 1;

while (v <= n) {

printf("%d\n", v);

v++;

}

​

return 0;

}

18

Number to count up to: 5

1

2

3

4

5

Result:

What happens if we try to combine the increment operator (++) with the condition?

Repeat N Times

PROBLEM: Given N, repeat an action N times

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SOLUTION: We need a counter variable that counts up from 0 to N-1 (idiomatic) or 1 to N (less common)

​

Using while:

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int i = 0; // counter variable

while (i < n) {

// Perform action ...

i++;

}

However, it is more common to use the for loop, which we will discuss next week

​

Using for:

​

for (int i = 0; i < n; i++) {

// Perform action ...

}

�Counter variables are commonly named i, j, or so, but you’re welcome to use a more meaningful name

19

CODING PATTERN

PATTERN NAME

++ and -- can make your head spin

• Given a = 10 and b = 20, what are the final values of a, b, and c in the following statements?
• c = a++ + b++;
• c = 10 + ++a - b--;
• c = ++a * 2 + ++a * 3;
• The last example is BAD – the result depends on the order of executing the two ++ operators, which is undefined
• Never use these operators on the same variable more than once in the same expression (you’d probably never have a reason to do it anyway)

20

INTERESTING BITS

Problem solving recipe: loop invariants

Problem: Sum of the inputs

• Given an integer n, followed by exactly n integers, find the sum of the n integers
• Example:

5

1 3 2 -1 7

~> 12

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• We cannot add up all the integers at once – instead we need to go through them one-by-one
• We need a variable to hold the result as we progress
• We can use Repeat N Times pattern to read and process the n inputs

22

Solution: Sum of the inputs

#include <stdio.h>

​

int main() {

int n;

int num;

int sum = 0;

​

// Read the number of inputs

scanf("%d", &n);

int i = 0;

while (i < n) {

scanf("%d", &num); // read each number

sum += num; // add to the sum

i++;

}

printf("%d\n", sum);

return 0;

}

23

Note the variable sum:

• At any point, sum remains the sum of the inputs so far
• Before the loop starts, sum is the sum of zero inputs, which is exactly zero
• At the end of each iteration, sum is the sum of the inputs read so far
• When the loop finishes, sum is the sum of all inputs – the final and expected result

Loop invariants

• In the previous example, the condition that “the variable sum must remain the sum of all inputs so far” is called the loop invariant
• Invariant means “an unchanging property”

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• A loop invariant is a condition or statement about a loop's variables that is true before the loop starts and remains true after every single iteration

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• Loop invariants are especially useful in problems that deal with directly deriving a single result from a list of items

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• By maintaining the invariant, correct result is guaranteed by the end of the loop

24

A simple 4-step recipe to use loop invariants

1. Result Variable: Set up a variable that will hold your final result (e.g., max_val, sum, count)
2. Initialization: Initialize this variable with a base case value that makes your rule true even before the loop starts
• Example: For summing, start with 0. For finding the max, start with the first item.
3. Maintenance: Inside the loop, update the variable as you process each new item, ensuring your rule (invariant) is still true at the end of every iteration
• The Invariant: "My variable holds the correct result for all the items I've seen so far."
4. Termination: When the loop finishes, the variable will hold the final, correct result

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• This recipe works well for problems with a single-value result that can be derived directly from a list of items
• However, this is not the only way to use loop invariants, but it’s a simple way to start

25

Problem: Maximum of the inputs

• Given an integer n, followed by exactly n integers, find the maximum of the n integers

​

• Let’s use the simple loop invariant recipe:
• Result Variable: The result is the maximum, so we name it max_val
• Let’s pause here: What’s the loop invariant for this problem?
• Invariant: max_val holds the maximum of all the inputs we’ve seen so far
• Initialization: Maximum of zero inputs is undefined, so we initialize max_val to the first item as its base case value instead
• Maintenance: Inside the loop, update max_val such that it always holds the maximum of the inputs seen so far by the end of each iteration
• Termination: When the loop finishes, max_val should hold the maximum of all the inputs – verify that it does

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• Work it out on the provided blank sheet to go from the recipe to code

26

Blank sheet – work out your solutions here

27

Problem: Factorial of N

• Given an integer n, 0 ≤ n ≤ 10, compute n!

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• Again, let’s use the simple loop invariant recipe to work it out:
• What’s the result variable?
• What’s the invariant?
• What’s the initial value?
• How do we maintain the invariant?

​

• Work it out on the provided blank sheet

28

Blank sheet – work out your solutions here

29

Solution: Factorial of 10

#include <stdio.h>

​

int main()

{

int n = 10;

int fact = 1;

​

int i = 1;

while (i <= n) {

fact *= i;

i++;

}

​

printf("%d! = %d\n", n, fact);

​

return 0;

}

30

10! = 3628800

Result:

Problem: Sum up until 50

• Find the largest N such that the sum from 1 to N is still less than 50
• This problem is similar to sum of inputs, but with a different terminating condition
• It’s not counting up to N, but to add up to some value less than 50
• The terminating condition has 50 as its end marker

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• In this problem, it’s not exactly the sum that we want to know – it’s the N that sums up to it
• Can we use the simple loop invariant recipe to solve it?
• What’s the result variable?
• What’s the invariant?

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• Loop invariants sometimes involve more than one variables – such as in this problem – so the simple loop invariant recipe does not apply
• As you gain more experience with simpler loop problems, this one will also become easier

31

Solution: Sum up until 50

#include <stdio.h>

​

int main()

{

int target = 50;

int n = 1, sum = 0;

​

while (sum + n < target) {

sum += n;

n++;

}

​

printf("N that sums up to %d = %d\n", sum, n - 1);

​

return 0;

}

32

N that sums up to 45 = 9

Result:

while (sum + n < target)

sum += n++;

More concise version

Interrupting the loop flow

break and continue

• These statements are used to interrupt the flow of loops

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• break exits from the loop
• continue skips the remaining statements and starts the next iteration

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• Often used with a conditional statement (if or if … else)

34

Sum of a series with skipping

#include <stdio.h>

​

int main()

{

int n = 0, sum = 0, last_add = 0;

​

while (1) { // infinite loop

n++;

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// Skip any number divisible by 3

if (n % 3 == 0)

continue;

​

// Stop once target is reached

if (sum + n >= 35)

break;

​

sum += n;

last_add = n; // we record last_add because sometimes n is skipped

}

​

printf("N that sums up to %d = %d\n", sum, last_add);

return 0;

}

35

N that sums up to 27 = 8

Result:

We have just seen an infinite loop

• A 1 means true, which, in this case, means the condition is always true -- indefinitely

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• We use an infinite loop when:
• The exit condition can only be checked in the middle of the iteration
• There are multiple exit conditions
• The task really needs to be executed indefinitely
• For examples, game or event-handling loops

36

The do … while loop

do … while loop – act first, then check

• Next is do … while loop
• Syntax: do statement while (condition);

​

• The do … while loop repeatedly executes statement until condition is false
• The statement is executed first, then the condition is checked – at least one iteration is guaranteed

38

Condition

Statement(s)

False

True

Problem: Input validation

• Problem: We want to ask the user to guess a number, but also make sure the input is within the range 1-100
• Solution: If the user enters a value that is outside the range, keep asking until a valid input is entered

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• This is part of defensive programming practice – we assume that the user always makes mistakes

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• Input validation is very common:
• Asking the user to confirm (with ‘y’ or ‘n’)
• Presenting a list of choices and making sure the user enters a valid one

39

Input validation

#include <stdio.h>

​

int main()

{

int guess;

​

do {

printf("Enter your guess (1-100): ");

scanf("%d", &guess);

} while (guess < 1 || guess > 100);

​

printf("You guessed %d.\n", guess);

​

return 0;

}

40

Enter your guess (1-100): 0

Enter your guess (1-100): 101

Enter your guess (1-100): 100

You guessed 100.

Result:

Can we use while loop for that?

#include <stdio.h>

​

int main()

{

int guess;

​

printf("Enter your guess (1-100): ");

scanf("%d", &guess);

while (guess < 1 || guess > 100) {

printf("Enter your guess (1-100): ");

scanf("%d", &guess);

}

​

printf("You guessed %d.\n", guess);

​

return 0;

}

41

Code is repeated

When to use which?

• Use while in most cases for conditional loops

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• Use do … while when one of the followings holds:
• You need to always have at least one iteration regardless of the condition
• The value you’re checking requires the loop body to run to get it

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• Input validation is the most common use case for do … while

42

When you’re lost, a reference is here to help…

C reference: https://en.cppreference.com/w/c

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Still, I recommend getting a book – your life will be easier that way

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That’s all for today!

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