Arrays, Strings, and Pointers

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
• L05 Arrays, strings, and pointers (today!)
• 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 for loop
• Arrays
• Strings
• Pointers
• Pointer arithmetic

3

Previous lesson recaps

4

while loop – check first, then act

#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;

}

5

Condition

Statement(s)

False

True

Repeat N Times

PROBLEM: Given N, repeat an action N times

​

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

​

Using while:

​

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

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CODING PATTERN

PATTERN NAME

Simple Loop Invariant

This is a simple 4-step recipe to use loop invariants that works well for problems with a single-value result that can be derived directly from a list of items

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

However, this is not the only way to use loop invariants, but it’s a simple way to start

7

CODING RECIPE

RECIPE NAME

break and continue

• These statements are used to interrupt the flow of loops

​

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

​

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

8

do … while loop – act first, then check

#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;

}

9

Condition

Statement(s)

False

True

The for loop

Convenient counter-controlled style for loop

​

• This while loop pattern is used often

​

initialize;

while (condition) {

statement(s);

step;

}

​

• But it’s more concise to use for loop for this pattern

​

for (initialize; condition; step)

statement

11

1

2

4

3

1

2

4

3

True

False

Factorial of 10 (for loop version)

#include <stdio.h>

​

int main()

{

int n = 10;

int fact = 1;

​

for (int i = 1; i <= n; i++)

fact *= i;

​

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

​

return 0;

}

12

int i = 1;

while (i <= n) {

fact *= i;

i++;

}

Compared with

Scope of the counter variable

• The for loop here:

​

for (int i = 1; i <= n; i++)

fact *= i;

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

​

• There is a compilation error – variables declared inside the for loop are not available outside it
• ͏Scope is the region of code where a variable is accessible
• In this case, i has a local scope inside the for loop only

• If we move the declaration out:

​

int i;

for (i = 1; i <= n; i++)

fact *= i;

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

​

• It will now compile, because the variable is declared outside the for loop
• Here, the scope of i is outside the loop, so it is now accessible to printf()

13

When to use which?

• Prefer while for conditional loops in general

​

• Prefer for when you want a counter-controlled loop or when you’re working with a list of items

​

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

​

• These are general guidelines, not rules – use whatever that works best for your specific situations!

14

Arrays

Problem: Counting integers equal or above the average

• Given a list of n integers, count items whose value is equal to or above average
• Let’s U-SWACT it together!
• UNDERSTAND: Take a minute to understand the problem first
• What are the inputs?
• What is the expected output?
• What’s missing here? – we need to know the average before we can count!
• SAMPLE INPUTS: Create some sample inputs
• We need a few, but let’s try: 7 5 3 4 2
• WORK OUT: This is quite simple – try it!
• Remember to work through the list one at a time

Help! We’re stuck!

• ALGORITHM: So far so good, so we need to find an average then count the elements
• But how do we go back and count the elements we’ve read?
• We need something to hold the values of multiple elements!
• This is what we call an array

​

• There are a few things we need to solve first
• How do we read user inputs into an array?
• How do we sum an array so we can compute the average?
• How do we go through and check each element in the array if it’s qualified for a condition?

17

Arrays

• Array is a fixed-size collection of elements of the same type
• Each element is accessible via an index
• Declaration and initialization:

// 10-element int array (40 bytes allocated), uninitialized

int xs[10];

// 3-element float array (24 bytes allocated), with initialization

double ys[] = { 1.0, 2.5, 4.5 };

// 5-element unsigned int array (20 bytes), initialized to 1, 2, 3, 0, 0

unsigned zs[5] = { 1, 2, 3 };

18

Operations on arrays

• Array indexes starts from 0 to N-1
• Use an integer expression as a subscript (inside square brackets) to refer to an element

​

int arr[3];

​

arr[0] = 1;

arr[1] = arr[0] + 2;

arr[arr[0] + 1] = arr[0] + arr[1];

19

Reading user inputs into an array

• Using Repeat N Times, we perform scanf() for each element

​

int main()

{

int scores[10];

​

// Read user inputs into array

for (int i = 0; i < 10; i++)

scanf("%d", &scores[i]); // array elements need &

​

// Do other stuff ...

}

20

Sum of the array

#include <stdio.h>

​

int main()

{

int arr[] = { 3, 4, 2, 0, 1 };

int num_elems = 5;

int sum = 0;

​

for (int i = 0; i < num_elems; i++)

sum += arr[i];

​

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

​

return 0;

}

21

Sum = 10

Result:

Solution: Counting integers equal or above the average

int elems[MAX_ELEMS];

int num_elems;

​

scanf("%d", &num_elems);

​

int sum = 0;

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

scanf("%d", &elems[i]);

sum += elems[i];

}

​

float average = (float)sum / num_elems;

​

int count = 0;

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

if (elems[i] >= average) {

count++;

}

}

​

printf("Count of elements above %.2f: %d\n", average, count);

22

5

7 5 3 4 2

Count of elements above 4.20: 2

Result:

Mixing data types require casting, otherwise our answer will be wrong

Example: Finding the maximum

#define MAX_LEN 100

​

int main()

{

int scores[MAX_LEN]; // pre-allocated size must be large enough

int num_scores;

​

// Read number of inputs

scanf("%d", &num_scores);

// Read user inputs into array

for (int i = 0; i < num_scores; i++)

scanf("%d", &scores[i]); // array elements need &

​

// Find maximum score

int max = scores[0];

for (int i = 1; i < num_scores; i++)

if (scores[i] > max)

max = scores[i];

​

printf("Max = %d\n", max);

}

23

5�5 7 2 9 6

Max = 9

Result:

Example: Searching for an element

int elems[] = { 6, 1, 3, 0, 9, 4, 2 };

int target;

int pos = -1;

​

printf("Enter a number to search: ");

scanf("%d", &target);

​

for (int i = 0; i < 7; i++) {

if (elems[i] == target) {

pos = i;

break;

}

}

​

if (pos != -1) {

printf("Found %d at position %d\n", target, pos);

} else {

printf("%d not found in the array\n", target);

}

24

Enter a number to search: 9

Found 9 at position 4

​

// Re-run

Enter a number to search: 5

5 not found in the array

Result:

Memory addresses and pointers

Variables, memory, and addresses

26

• Variables are stored in memory
• Memory is a collection of byte-addressable cells

​

int exp = 100000; // 0x186A0

char name[8] = "Beck"; // 0x42 0x65 0x63 0x6B 0x00

int *ptr = &exp; // 0xBC04

​

• ͏Pointers are variables that store addresses – they point to data stored in memory
• Note that actual in-memory ordering of variables is not guaranteed to be the same as the declaration order

Pointers

• Pointer variable declaration: type *name;

int *ptr;

• “Address of” operator (&) is used to obtain the address of a variable

ptr = &exp;

• Array and string variable names already represent their memory addresses by themselves – we can get their addresses directly without using & operator just by referring their names

​

• “Pointer dereference” operator (*) is used to get the value stored at the memory address pointed to by the pointer

*ptr = *ptr + 1000;

• It is also used for assigning a value to that memory address

27

Pointers in action

int arr[] = { 3, 4, 2, 0, 1 };

int *ptr = arr;

​

for (int i = 0; i < 5; i++) {

printf("Value at %p = %d\n", ptr, *ptr);

ptr++;

}

printf("Address of the pointer = %p\n", &ptr);

​

• Use %p to print the address referred by the pointer
• Pointer moves (++, --) in a step size equal to the size of the referred type (4 bytes for int in this case)
• Pointer is also a variable, and has its own memory address, too!

28

Value at 00000000005FFE80 = 3

Value at 00000000005FFE84 = 4

Value at 00000000005FFE88 = 2

Value at 00000000005FFE8C = 0

Value at 00000000005FFE90 = 1

Address of the pointer = 00000000005FFE78

Result:

Strings

29

Greeting with “strings”

#include <stdio.h>

​

int main()

{

char name[50];

char surname[50];

int age;

​

printf("Enter your name: ");

scanf("%s", name);

printf("Enter your surname: ");

scanf("%49s", surname);

printf("Enter your age: ");

scanf("%d", &age);

​

printf("Hi %s %s, you are %d years old!\n", name, surname, age);

}

30

Enter your name: John

Enter your surname: Doe

Enter your age: 25

Hi John Doe, you are 25 years old!

Result:

C strings are arrays of char

Use %s to read a string (word)

Note that string variables do not have an & in front of them

Also, use %s to print strings

String reading can be length-limited

C uses null-terminated strings

• Strings are sequence of characters, used for representing textual data

​

• C-style strings are null-terminated char arrays
• Null-terminated means that strings end with the null character ('\0') which is an integer with the value 0

​

• Null-terminated strings are not used in most other languages
• C++ and Java which have a native string type, although C++ also supports null-terminated strings

31

Null-terminated strings

char text[10] = "Hello!";

​

• This example is a string pre-allocated to 10 bytes, using 7 bytes to store a string of length 6

32

Reading strings with scanf()

• When used with string (%s), scanf() will read characters until a whitespace is encountered

char str[20];

printf("Enter a word: ");

scanf("%s", str);

​

• If “hello world” is entered, only “hello” will be read into str
• No need to use &str, the array name (str) already represents its address

​

• To prevent buffer overflow, width specifier can be used

scanf("%19s", word);

• Note that the width does not include the null character – therefore we specify 1 character less than the array size

33

Problem: Counting string length

• Given a string, find the length of the string

​

• How do we determine a string length?
• By counting characters?
• When will we finish counting?

​

• Think again: What is the last character in every string?
• It’s the null character ('\0’)

​

• Can we use the Simple Loop Invariant recipe?
• Let’s try!

• Result Variable: We name it count
• Invariant: count is always the length of the string we’ve seen so far
• Initialization: Before the loop, we haven’t seen any character yet, so we initialize count to zero
• Maintenance: At the end of each iteration, we’ve seen one more character, so we count up by one
• Termination: How (and when) do we terminate?
• Terminating condition is different from the Repeat N Times pattern!

​

• So, how should we end this loop?
• What’s the terminating condition?

34

Solution: Counting string length

#include <stdio.h>

​

int main()

{

char str[20];

​

printf("Enter a word: ");

scanf("%19s", str);

� int count = 0;

while (str[count] != '\0')

count++;

� printf("Length of \"%s\" = %d\n", str, count);

​

return 0;

}

35

Enter a word: Hello!

Length of "Hello!" = 6

Result

Convenient function for string length

• Getting the length of a string is a very common operation
• C provides a library function for this operation in string.h
• The function is strlen(s)
• You’ll need to #include <string.h> to use it

​

• Common usage pattern:

​

for (int i = 0; i < strlen(s); i++) {

// Do something

}

36

Problem: Find and replace characters

• Given a string s, a character c we want to look for, and a character d to replace it, find all occurrences of c in s, and replace them with d

​

• We now know how to loop from the first character to the last

37

Solution: Find and replace characters

char str[50];

char src;

char dest;

​

printf("Enter a string: ");

fgets(str, sizeof(str), stdin);

printf("Enter character to replace: ");

scanf(" %c", &src);

printf("Enter replacement character: ");

scanf(" %c", &dest);

​

int i = 0;

while (str[i] != '\0') {

if (str[i] == src) {

str[i] = dest;

}

i++;

}

​

printf("Modified string: %s\n", str)

38

for (int i = 0; str[i] != '\0'; i++) {

if (str[i] == src) {

str[i] = dest;

}

}

With for loop

That’s all for today!

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