Pointer and variable relationship in C++ - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a pointer to an integer variable.

C++

int x = 10;
int [1] ptr = &x;

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

A*

B&

C#

D%

Attempts:

3 left

💡 Hint

Common Mistakes

Using & instead of * when declaring a pointer.

Confusing pointer declaration with address-of operator.

The asterisk (*) is used to declare a pointer variable in C++.

2fill in blank

medium

Complete the code to assign the address of variable x to pointer ptr.

C++

int x = 5;
int* ptr;
ptr = [1]x;

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

A*

B&&

C%

D&

Attempts:

3 left

💡 Hint

Common Mistakes

Using * instead of & when assigning address.

Using && which is for rvalue references, not addresses.

The ampersand (&) operator gives the address of a variable.

3fill in blank

hard

Fix the error in the code to correctly print the value pointed to by ptr.

C++

int x = 20;
int* ptr = &x;
std::cout << [1]ptr << std::endl;

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

A&ptr

Bptr

C*ptr

D&&ptr

Attempts:

3 left

💡 Hint

Common Mistakes

Printing the pointer itself instead of the value it points to.

Using &ptr which gives the address of the pointer variable, not the value.

Using *ptr dereferences the pointer to get the value it points to.

4fill in blank

hard

Fill both blanks to declare a pointer and assign it the address of variable y.

C++

int y = 15;
[1] ptr = [2]y;

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

Aint*

Bint

C&

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Using int instead of int* for pointer declaration.

Using *y instead of &y when assigning address.

int* declares a pointer to int, and &y gets the address of y.

5fill in blank

hard

Fill all three blanks to declare a pointer, assign it the address of z, and print the value pointed to by ptr.

C++

int z = 30;
[1] ptr = [2]z;
std::cout << [3]ptr << std::endl;

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

Aint*

B&

C*

Dint

Attempts:

3 left

💡 Hint

Common Mistakes

Using int instead of int* for pointer declaration.

Assigning *z instead of &z.

Printing ptr instead of *ptr.

int* declares the pointer, &z gets the address, and *ptr dereferences to the value.

Practice

1. What does a pointer variable in C++ store?

easy

A. The value of another variable

B. The memory address of another variable

C. The size of a variable

D. The type of a variable

Solution

1. Step 1: Understand what a pointer is

   A pointer is a variable that stores the address of another variable, not the value itself.

2. Step 2: Differentiate pointer from normal variables

   Normal variables store values directly, but pointers store where those values are located in memory.

3. Final Answer:

   The memory address of another variable -> Option B

4. Quick Check:

   Pointer = address storage [OK]

Hint: Pointers always hold addresses, not values [OK]

Common Mistakes:

• Confusing pointer value with variable value
• Thinking pointer stores variable type
• Assuming pointer stores variable size

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. ptr int*;

Solution

1. Step 1: Recall pointer declaration syntax

   In C++, to declare a pointer to an int, you write the type followed by an asterisk and the pointer name.

2. Step 2: Check each option

   Only 'int *ptr;' matches the correct syntax. Others have misplaced asterisks or wrong order.

3. Final Answer:

   int *ptr; -> Option C

4. Quick Check:

   Pointer declaration = type *name [OK]

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

Common Mistakes:

• Placing * after variable name
• Using * before type keyword
• Mixing order of type and pointer symbol

3. What is the output of the following code?

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

medium

A. 20

B. 10

C. Address of x

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, so x becomes 20.

2. Step 2: Output the value of x

   After changing *p, printing x outputs 20.

3. Final Answer:

   20 -> Option A

4. Quick Check:

   Pointer dereference changes variable value [OK]

Hint: Changing *p changes x's value [OK]

Common Mistakes:

• Thinking *p = 20 changes pointer, not value
• Expecting output to be address
• Confusing pointer and variable values

4. Identify the error in the following code:

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

medium

A. Pointer p is not initialized before dereferencing

B. Variable x is not declared

C. Incorrect pointer declaration syntax

D. Missing semicolon after declaration

Solution

1. Step 1: Check pointer initialization

   Pointer p is declared but not assigned an address before using *p = x; which causes undefined behavior.

2. Step 2: Understand dereferencing uninitialized pointer

   Dereferencing p without initialization leads to runtime error or crash.

3. Final Answer:

   Pointer p is not initialized before dereferencing -> Option A

4. Quick Check:

   Always initialize pointers before dereference [OK]

Hint: Initialize pointers before use to avoid errors [OK]

Common Mistakes:

• Dereferencing uninitialized pointers
• Assuming pointer auto-initializes
• Ignoring runtime errors from bad pointers

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. 3 4

B. 3 3

C. 4 4

D. 4 3

Solution

1. Step 1: Understand pointer swapping

   Initially, p points to a (3), q points to b (4). temp stores p (address of a).

2. Step 2: Swap pointers

   p = q makes p point to b (4), q = temp makes q point to a (3).

3. Step 3: Print values pointed by p and q

   *p is 4, *q is 3, so output is "4 3".

4. Final Answer:

   4 3 -> Option D

5. Quick Check:

   Pointer swap changes what they point to [OK]

Hint: Swapping pointers swaps pointed values [OK]

Common Mistakes:

• Confusing pointer values with pointed values
• Mixing up which pointer points where after swap
• Assuming values themselves swap