Pointer declaration - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of pointer declaration and usage

What is the output of this C code snippet?

C

#include <stdio.h>

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

A Address of x

B 0

C 10

D Compilation error

💡 Hint

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

The pointer p stores the address of x. Using *p accesses the value of x, which is 10.

❓ Predict Output

intermediate

Value of pointer variable after declaration

What is the value of the pointer variable p after this code runs?

C

int x = 5;
int *p = &x;

A Undefined value

B The address of x

C The value 5

D NULL

💡 Hint

Pointers store addresses, not values directly.

The pointer p is assigned the address of x using &x, so p holds x's address.

📝 Syntax

advanced

Identify the correct pointer declaration

Which of the following is a correct declaration of a pointer to an integer?

A int *p;

B int p*;

C int &p;

D pointer int p;

💡 Hint

The asterisk * goes before the pointer name to declare a pointer.

In C, 'int *p;' declares p as a pointer to an int. Other options are invalid syntax.

🧠 Conceptual

advanced

Pointer declaration with multiple variables

What is the type of variables a and b after this declaration? int *a, b;

A a is a pointer to int; b is an int

B a and b are both pointers to int

C a and b are both ints

D a is an int; b is a pointer to int

💡 Hint

The asterisk applies only to the variable it precedes.

Only a is declared as a pointer to int. b is a normal int variable.

🔧 Debug

expert

Find the error in pointer declaration

What error does this code produce? int *p; *p = 10;

C

int *p;
*p = 10;

A No error, prints 10

B Compilation error: missing semicolon

C Type error: incompatible types

D Segmentation fault or undefined behavior

💡 Hint

p is declared but not initialized before dereferencing.

Pointer p is uninitialized and points to an unknown location. Writing *p = 10 causes undefined behavior or crash.

Practice

1. What does the following declaration mean in C?
int *ptr;

easy

A. ptr is a pointer to a character variable

B. ptr is a pointer to an integer variable

C. ptr is an integer variable

D. ptr is a function returning an integer

Solution

1. Step 1: Understand the * symbol in declaration

   The * before ptr means ptr is a pointer variable, not a normal integer.

2. Step 2: Check the type before *

   The type before * is int, so ptr points to an integer variable.

3. Final Answer:

   ptr is a pointer to an integer variable -> Option B

4. Quick Check:

   int *ptr means pointer to int [OK]

Hint: Look for * before variable name to identify pointer [OK]

Common Mistakes:

• Thinking * means multiplication here
• Confusing pointer with normal variable
• Ignoring the type before *

2. Which of the following is the correct way to declare a pointer to a float variable in C?

easy

A. ptr float*;

B. float ptr*;

C. *float ptr;

D. float *ptr;

Solution

1. Step 1: Recall pointer declaration syntax

   Pointer declaration syntax is: type *name; so the * goes between type and variable name.

2. Step 2: Check each option

   float *ptr; matches correct syntax: float *ptr; others have wrong order or missing parts.

3. Final Answer:

   float *ptr; -> Option D

4. Quick Check:

   Pointer syntax = type *name; [OK]

Hint: Always put * between type and variable name [OK]

Common Mistakes:

• Placing * after variable name
• Putting * before type
• Missing * entirely

3. What is the output of this C code?

int x = 10;
int *p = &x;
printf("%d", *p);

medium

A. 10

B. Compilation error

C. Garbage value

D. Address of x

Solution

1. Step 1: Understand pointer initialization

   p is assigned the address of x using &x, so p points to x.

2. Step 2: Dereference pointer to get value

   *p means value at address p, which is x's value 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   *p gives value at pointer address [OK]

Hint: Use * to get value from pointer address [OK]

Common Mistakes:

• Printing pointer p instead of *p
• Confusing address and value
• Forgetting & to get address

4. Identify the error in this pointer declaration:

int *p;
p = 5;

medium

A. Error: assigning integer directly to pointer

B. Correct: p points to integer 5

C. Error: missing * in assignment

D. Error: pointer must be initialized with & operator

Solution

1. Step 1: Check pointer assignment rules

   A pointer must hold a memory address, not a direct integer value.

2. Step 2: Analyze given assignment

   Assigning 5 directly to pointer p is invalid; it should be an address like &variable.

3. Final Answer:

   Error: assigning integer directly to pointer -> Option A

4. Quick Check:

   Pointer = address, not direct value [OK]

Hint: Pointers store addresses, never assign raw values directly [OK]

Common Mistakes:

• Assigning integer instead of address
• Forgetting & operator for address
• Confusing pointer and normal variable

5. Given the code below, which pointer declaration correctly allows modifying the value of num through the pointer?

int num = 20;
// Which declaration is correct?

hard

A. int *ptr = num;

B. int ptr = #

C. int *ptr = #

D. int &ptr = num;

Solution

1. Step 1: Understand pointer declaration for modifying variable

   To modify num via pointer, pointer must hold address of num, declared as int *ptr.

2. Step 2: Check each option

   int *ptr = # correctly declares pointer to int and assigns address of num. Others are invalid syntax or wrong types.

3. Final Answer:

   int *ptr = # -> Option C

4. Quick Check:

   Pointer to int = &variable for modification [OK]

Hint: Use int *ptr = &var to modify var via pointer [OK]

Common Mistakes:

• Assigning value instead of address
• Using wrong type for pointer
• Confusing references with pointers