Address and dereference operators - Mini Project: Build & Apply

Address and Dereference Operators in C

📖 Scenario: You are working with a simple program that stores a number and uses pointers to access and modify it. This is like having a remote control (pointer) to change the TV (variable) without touching it directly.

🎯 Goal: Learn how to use the & operator to get the address of a variable and the * operator to access or change the value at that address.

📋 What You'll Learn

Create an integer variable named number with the value 10.

Create a pointer variable named ptr that stores the address of number.

Use the pointer ptr to change the value of number to 20.

Print the value of number after modification.

💡 Why This Matters

🌍 Real World

Pointers let programs work with memory directly, which is important for efficient software like games, operating systems, and device drivers.

💼 Career

Understanding pointers is essential for C programming jobs, embedded systems development, and performance-critical applications.

1

Create an integer variable

Create an integer variable called number and set it to 10.

C

// Create an integer variable called number and set it to 10
// Your code here

Hint

Use int number = 10; to create the variable.

2

Create a pointer to the variable

Create a pointer variable called ptr that stores the address of number using the & operator.

C

int number = 10;
// Create a pointer variable ptr that stores the address of number
// Your code here

Hint

Use int *ptr = &number; to create the pointer.

3

Change the value using the pointer

Use the pointer ptr and the dereference operator * to change the value of number to 20.

C

int number = 10;
int *ptr = &number;
// Change the value of number to 20 using ptr
// Your code here

Hint

Use *ptr = 20; to change the value at the address.

4

Print the modified value

Print the value of number using printf to show it has changed to 20.

C

int number = 10;
int *ptr = &number;
*ptr = 20;
// Print the value of number
// Your code here

Hint

Use printf("%d\n", number); to print the value.

Practice

1. What does the & operator do in C?

easy

A. It gives the memory address of a variable.

B. It multiplies two numbers.

C. It dereferences a pointer to get the value.

D. It declares a pointer variable.

Solution

1. Step 1: Understand the & operator

   The & operator is used to get the address of a variable in memory.

2. Step 2: Differentiate from other operators

   It does not multiply, dereference, or declare pointers; those are different operations.

3. Final Answer:

   It gives the memory address of a variable. -> Option A

4. Quick Check:

   & means address-of [OK]

Hint: Remember & means 'address of' variable [OK]

Common Mistakes:

• Confusing & with multiplication operator
• Thinking & dereferences a pointer
• Mixing & with pointer declaration

2. Which of the following is the correct way to declare a pointer to an integer in C?

easy

A. *int ptr;

B. int &ptr;

C. int ptr*;

D. int *ptr;

Solution

1. Step 1: Recall pointer declaration syntax

   In C, a pointer to an int is declared as int *ptr; where * indicates pointer type.

2. Step 2: Check other options for syntax errors

   int &ptr; is invalid in C (it's C++ reference), int ptr*; and *int ptr; are incorrect syntax.

3. Final Answer:

   int *ptr; -> Option D

4. Quick Check:

   Pointer declaration uses * before variable name [OK]

Hint: Pointer declaration: type *name; [OK]

Common Mistakes:

• Using & instead of * for pointers
• Placing * after variable name incorrectly
• Confusing C++ references with pointers

3. What is the output of this code?

int x = 10;
int *p = &x;
printf("%d", *p);

medium

A. 10

B. The memory address of x

C. Garbage value

D. Compilation error

Solution

1. Step 1: Understand pointer assignment

   Pointer p stores the address of x using &x.

2. Step 2: Dereference pointer to get value

   *p accesses the value at the address stored in p, which is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   *p gives value at address [OK]

Hint: *p gives value stored at pointer address [OK]

Common Mistakes:

• Printing pointer variable instead of dereferenced value
• Confusing & and * operators
• Expecting address instead of value

4. Identify the error in this code snippet:

int x = 5;
int *p;
*p = x;
printf("%d", *p);

medium

A. Incorrect use of & operator.

B. Pointer p is not initialized before dereferencing.

C. Variable x is not declared.

D. Missing semicolon after declaration.

Solution

1. Step 1: Check pointer initialization

   Pointer p is declared but not assigned an address before *p = x;.

2. Step 2: Understand dereferencing uninitialized pointer

   Dereferencing p without initialization causes undefined behavior or crash.

3. Final Answer:

   Pointer p is not initialized before dereferencing. -> Option B

4. Quick Check:

   Always assign pointer before dereferencing [OK]

Hint: Initialize pointer before using * to access value [OK]

Common Mistakes:

• Dereferencing uninitialized pointer
• Confusing assignment with address-of
• Ignoring pointer initialization warnings

5. Given the code below, what will be the output?

int a = 3, b = 7;
int *p = &a;
*p = *p + 4;
p = &b;
*p = *p - 2;
printf("%d %d", a, b);

hard

A. 7 3

B. 3 5

C. 7 5

D. 7 7

Solution

1. Step 1: Modify value pointed by p initially

   p points to a. *p = *p + 4; changes a from 3 to 7.

2. Step 2: Change pointer to point to b and modify

   p = &b; points to b. *p = *p - 2; changes b from 7 to 5.

3. Step 3: Print final values

   a is 7 and b is 5.

4. Final Answer:

   7 5 -> Option C

5. Quick Check:

   Pointer changes values via dereference [OK]

Hint: Track pointer target and value changes stepwise [OK]

Common Mistakes:

• Not updating the correct variable via pointer
• Confusing pointer assignment with value assignment
• Ignoring pointer reassignment effects