Pointer arithmetic in C++ - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer arithmetic with arrays

What is the output of the following C++ code?

C++

#include <iostream>
int main() {
    int arr[] = {10, 20, 30, 40};
    int* p = arr;
    std::cout << *(p + 2) << std::endl;
    return 0;
}

A 10

B 30

C 20

D 40

💡 Hint

Remember that pointer arithmetic moves by the size of the data type.

The pointer p points to the first element of arr (10). Adding 2 moves it two integers ahead, pointing to 30.

❓ Predict Output

intermediate

Pointer subtraction result

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

C++

#include <iostream>
int main() {
    int arr[] = {5, 10, 15, 20, 25};
    int* p1 = &arr[4];
    int* p2 = &arr[1];
    std::cout << p1 - p2 << std::endl;
    return 0;
}

A 6

B 4

C 5

D 3

💡 Hint

Pointer subtraction gives the number of elements between two pointers.

p1 points to arr[4], p2 to arr[1]. The difference is 4 - 1 = 3, but pointer subtraction counts elements, so output is 3.

❓ Predict Output

advanced

Pointer arithmetic with char arrays

What does this program print?

C++

#include <iostream>
int main() {
    char str[] = "hello";
    char* p = str;
    std::cout << *(p + 4) << std::endl;
    return 0;
}

A o

B h

C e

D l

💡 Hint

Pointer arithmetic on char arrays moves one character at a time.

The pointer p points to 'h'. Adding 4 moves to the fifth character 'o'.

❓ Predict Output

advanced

Pointer arithmetic with void pointers

What error does this code produce?

C++

#include <iostream>
int main() {
    int x = 10;
    void* p = &x;
    p = p + 1;
    return 0;
}

A Compilation error: invalid operands to binary expression

B Runtime error: segmentation fault

C No error, compiles and runs fine

D Compilation error: void pointer cannot be assigned

💡 Hint

Pointer arithmetic requires knowing the size of the pointed type.

You cannot do arithmetic on void pointers because the compiler doesn't know the size of the data type.

🧠 Conceptual

expert

Pointer arithmetic and array indexing equivalence

Which option correctly explains why *(arr + i) is equivalent to arr[i] in C++?

A Because arr[i] is a macro that expands to *(arr + i).

B Because arr[i] uses pointer subtraction internally to access elements.

C Because arr[i] is defined as *(arr + i) by the language standard.

D Because arr[i] is a function call that returns *(arr + i).

💡 Hint

Think about how array indexing is defined in terms of pointers.

The C++ standard defines arr[i] as *(arr + i), so they are exactly the same operation.