Address and dereference operators

Introduction

We use address and dereference operators to work with memory locations directly. This helps us access or change the value stored in a specific place in the computer's memory.

When you want to find the memory address of a variable.

When you want to access or change the value stored at a specific memory address.

When working with pointers to handle dynamic data or arrays.

When passing large data to functions efficiently by passing addresses instead of copies.

Syntax

C

int *pointer;  // declare a pointer to int
pointer = &variable;  // & is the address operator
value = *pointer;  // * is the dereference operator

The & operator gives the address of a variable.

The * operator accesses the value stored at the address the pointer holds.

Examples

Here, p stores the address of x. Using *p gets the value of x.

C

int x = 10;
int *p = &x;  // p holds the address of x
int y = *p;   // y gets the value 10 from x

Using *ptr = 20; changes the value stored at the address ptr points to, which is a.

C

int a = 5;
int *ptr = &a;
*ptr = 20;  // changes value of a to 20

Sample Program

This program shows how to get the address of a variable using &, access its value using *, and change the value through the pointer.

C

#include <stdio.h>

int main() {
    int num = 7;
    int *p = &num;  // p stores address of num

    printf("Value of num: %d\n", num);
    printf("Address of num: %p\n", (void*)&num);
    printf("Value stored at pointer p: %d\n", *p);

    *p = 15;  // change value of num using pointer
    printf("New value of num after change: %d\n", num);

    return 0;
}

Important Notes

Always initialize pointers before dereferencing to avoid errors.

Use (void*) cast when printing addresses with %p for portability.

Summary

The & operator gets the address of a variable.

The * operator accesses or changes the value at a given address.

Pointers store addresses and allow indirect access to variables.

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