Address and dereference operators in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the address of variable num.

C++

int num = 10;
int* ptr = [1]num;

Drag options to blanks, or click blank then click option'

A&

B*

C#

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using the dereference operator * instead of the address-of operator &.

Using symbols like # or % which are not related to pointers.

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

2fill in blank

medium

Complete the code to access the value pointed to by ptr.

C++

int num = 20;
int* ptr = #
int value = [1]ptr;

Drag options to blanks, or click blank then click option'

A&

B%

C*

D->

Attempts:

3 left

💡 Hint

Common Mistakes

Using the address-of operator & instead of dereference operator *.

Using the arrow operator -> which is for pointers to structs or classes.

The dereference operator * is used to access the value stored at the memory address a pointer holds.

3fill in blank

hard

Fix the error in the code to correctly assign the value 50 to the variable pointed to by ptr.

C++

int num = 0;
int* ptr = #
[1]ptr = 50;

Drag options to blanks, or click blank then click option'

A*ptr

B&ptr

Cptr

Dptr*

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning directly to the pointer variable instead of the value it points to.

Using incorrect syntax like ptr* or &ptr.

To assign a value to the variable pointed to by a pointer, use the dereference operator * before the pointer name.

4fill in blank

hard

Fill both blanks to create a pointer ptr that points to num and then access its value.

C++

int num = 100;
int* ptr = [1]num;
int val = [2]ptr;

Drag options to blanks, or click blank then click option'

A&

B*

C->

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping the operators in the blanks.

Using the arrow operator -> which is not for simple pointers.

Use & to get the address of num and * to access the value pointed to by ptr.

5fill in blank

hard

Fill all three blanks to declare an integer num, a pointer ptr to num, and assign the value 75 to num using the pointer.

C++

int [1];
int* [2] = &[3];
*ptr = 75;

Drag options to blanks, or click blank then click option'

Anum

Bptr

Dval

Attempts:

3 left

💡 Hint

Common Mistakes

Using different names for the variable and pointer inconsistently.

Not using the address-of operator & in the pointer initialization.

Declare num as the integer variable, ptr as a pointer to num, and use &num to get the address.

Practice

1. What does the & operator do in C++?

easy

A. It dereferences a pointer to get the value.

B. It multiplies two numbers.

C. It declares a pointer variable.

D. It gives the memory address of a 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, declare pointers, or dereference pointers; those are different operations.

3. Final Answer:

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

4. Quick Check:

   & operator = address retrieval [OK]

Hint: Remember & means 'address of' in C++ [OK]

Common Mistakes:

• Confusing & with multiplication operator
• Thinking & declares a pointer
• Mixing & with dereference * operator

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; declares a reference, not a pointer. int ptr*; and *int ptr; are invalid syntax.

3. Final Answer:

   int *ptr; -> Option A

4. Quick Check:

   Pointer declaration = int *ptr; [OK]

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

Common Mistakes:

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

3. What is the output of the following code?

int x = 10;
int *p = &x;
*p = 20;
std::cout << x << std::endl;

medium

A. Address of x

B. 10

C. 20

D. Compilation error

Solution

1. Step 1: Understand pointer assignment

   Pointer p stores the address of x. Using *p = 20; changes the value at that address.

2. Step 2: Check the effect on x

   Since *p points to x, changing *p to 20 updates x to 20.

3. Final Answer:

   20 -> Option C

4. Quick Check:

   Dereferencing pointer changes original variable [OK]

Hint: Changing *p changes the value of x [OK]

Common Mistakes:

• Thinking *p is the address, not the value
• Expecting output to be original x (10)
• Confusing pointer variable with pointed value

4. Identify the error in this code snippet:

int x = 5;
int *p;
*p = x;
std::cout << *p << std::endl;

medium

A. Variable x is not declared.

B. Pointer p is not initialized before dereferencing.

C. Syntax error in pointer declaration.

D. Cannot assign value to *p.

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 initialize pointers before dereferencing [OK]

Hint: Initialize pointers before using *p [OK]

Common Mistakes:

• Dereferencing pointer without assigning address
• Assuming pointer auto-points to variable
• Confusing pointer declaration with initialization

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

int a = 3, b = 4;
int *p = &a;
int *q = &b;
int *temp = p;
p = q;
q = temp;
std::cout << *p << " " << *q << std::endl;

hard

A. 4 3

B. 3 4

C. 3 3

D. 4 4

Solution

1. Step 1: Understand pointer swapping

   Initially, p points to a (3), q points to b (4). The code swaps the pointers using temp.

2. Step 2: Check final pointer values

   After swapping, p points to b (4), and q points to a (3).

3. Final Answer:

   4 3 -> Option A

4. Quick Check:

   Pointer swap changes what they point to [OK]

Hint: Swapping pointers swaps the values they point to [OK]

Common Mistakes:

• Confusing swapping pointers with swapping values
• Expecting original order after swap
• Mixing pointer values and pointed values