Address and dereference operators - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer dereference after assignment

What is the output of this C code snippet?

C

#include <stdio.h>

int main() {
    int x = 10;
    int *p = &x;
    *p = 20;
    printf("%d", x);
    return 0;
}

A 20

B 10

C Address of x

D Compilation error

💡 Hint

Remember that *p accesses the value at the address stored in p.

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

❓ Predict Output

intermediate

Value printed by pointer to pointer

What value does this program print?

C

#include <stdio.h>

int main() {
    int a = 5;
    int *p = &a;
    int **pp = &p;
    printf("%d", **pp);
    return 0;
}

A Address of a

B Address of p

C 5

D Compilation error

💡 Hint

**pp means dereference twice: first to get p, then to get a's value.

pp points to p, which points to a. So **pp accesses the value of a, which is 5.

🔧 Debug

advanced

Identify the error in pointer assignment

What error does this code produce when compiled?

C

#include <stdio.h>

int main() {
    int x = 10;
    int *p;
    *p = x;
    printf("%d", *p);
    return 0;
}

A Prints 10

B Compilation error: uninitialized pointer

C Prints garbage value

D Segmentation fault at runtime

💡 Hint

Consider what happens when you dereference an uninitialized pointer.

Pointer p is declared but not initialized to a valid address. Dereferencing it causes a runtime segmentation fault.

📝 Syntax

advanced

Correct pointer declaration syntax

Which option correctly declares a pointer to an integer variable?

A int *p;

B int &p;

C int p*;

D pointer int p;

💡 Hint

In C, the * symbol is used to declare a pointer.

Option A uses correct C syntax for declaring a pointer to int. Others are invalid syntax.

🚀 Application

expert

Predict the output of pointer arithmetic and dereference

What is the output of this program?

C

#include <stdio.h>

int main() {
    int arr[] = {10, 20, 30};
    int *p = arr;
    printf("%d", *(p + 2));
    return 0;
}

A 10

B 30

C 20

D Compilation error

💡 Hint

Pointer arithmetic moves the pointer by multiples of the data type size.

p points to arr[0]. Adding 2 moves to arr[2], which is 30. Dereferencing prints 30.