Pointer and variable relationship in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer and variable modification

What is the output of this C++ code snippet?

C++

#include <iostream>
int main() {
    int x = 10;
    int* p = &x;
    *p = 20;
    std::cout << x << std::endl;
    return 0;
}

A 20

B 10

C Compilation error

D Undefined behavior

💡 Hint

Remember that *p accesses the value pointed to by p.

The pointer p points to x. Changing *p changes x's value to 20. So printing x outputs 20.

❓ Predict Output

intermediate

Pointer reassignment effect

What is the output of this C++ code?

C++

#include <iostream>
int main() {
    int a = 5, b = 10;
    int* p = &a;
    p = &b;
    *p = 20;
    std::cout << a << " " << b << std::endl;
    return 0;
}

A 5 20

B 5 10

C 20 20

D 20 10

💡 Hint

Pointer p is changed to point to b before modifying *p.

Initially p points to a, but then p points to b. Changing *p changes b to 20. a remains 5.

❓ Predict Output

advanced

Pointer arithmetic and variable values

What is the output of this C++ code snippet?

C++

#include <iostream>
int main() {
    int arr[3] = {1, 2, 3};
    int* p = arr;
    *(p + 1) = 10;
    std::cout << arr[1] << std::endl;
    return 0;
}

A 2

B 10

C Compilation error

D Undefined behavior

💡 Hint

Pointer arithmetic moves the pointer to the next element.

p points to arr[0]. *(p + 1) accesses arr[1], which is set to 10. So arr[1] prints 10.

❓ Predict Output

advanced

Pointer to pointer value change

What is the output of this C++ program?

C++

#include <iostream>
int main() {
    int x = 5;
    int* p = &x;
    int** pp = &p;
    **pp = 15;
    std::cout << x << std::endl;
    return 0;
}

A 5

B Compilation error

C 15

D Undefined behavior

💡 Hint

**pp accesses the value pointed to by the pointer p.

pp points to p, which points to x. Changing **pp changes x to 15.

❓ Predict Output

expert

Pointer and variable relationship with const correctness

What is the output of this C++ code snippet?

C++

#include <iostream>
int main() {
    int x = 10;
    const int* p = &x;
    // *p = 20; // Uncommenting this line causes error
    x = 20;
    std::cout << *p << std::endl;
    return 0;
}

A Undefined behavior

B 10

C Compilation error

D 20

💡 Hint

const int* means you cannot change the value through the pointer, but the variable itself can change.

p points to x as a const pointer. Changing x directly to 20 updates the value. *p prints 20.

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