Pointer declaration in C++

Introduction

Pointers let you store the address of a variable. This helps you work directly with memory locations.

When you want to change a variable inside a function and see the change outside.

When you need to work with arrays or dynamic memory.

When you want to pass large data efficiently without copying.

When you want to create complex data structures like linked lists.

When you want to access hardware or system resources directly.

Syntax

C++

type *pointerName;

The asterisk (*) shows that the variable is a pointer.

The pointer stores the address of a variable of the specified type.

Examples

Declares a pointer p that can hold the address of an int variable.

C++

int *p;

Declares a pointer cptr that can hold the address of a char variable.

C++

char *cptr;

Declares a pointer dptr that can hold the address of a double variable.

C++

double *dptr;

Sample Program

This program shows how to declare a pointer and use it to access the value and address of a variable.

C++

#include <iostream>

int main() {
    int number = 42;
    int *ptr = &number;  // ptr stores the address of number

    std::cout << "Value of number: " << number << "\n";
    std::cout << "Address of number: " << &number << "\n";
    std::cout << "Value stored in ptr (address): " << ptr << "\n";
    std::cout << "Value pointed to by ptr: " << *ptr << "\n";

    return 0;
}

Important Notes

The actual address printed will be different each time you run the program.

Use the & operator to get the address of a variable.

Use the * operator to get the value stored at the address the pointer points to.

Summary

Pointers store memory addresses of variables.

Use type *name; to declare a pointer.

Use & to get an address and * to access the value at that address.

Practice

1. Which of the following correctly declares a pointer to an integer in C++?

easy

A. ptr int*;

B. int ptr*;

C. *int ptr;

D. int *ptr;

Solution

1. Step 1: Understand pointer declaration syntax

   In C++, a pointer to a type is declared by writing the type followed by an asterisk (*) and then the pointer name.

2. Step 2: Check each option

   int *ptr; uses int *ptr; which is the correct syntax. Other options place the asterisk incorrectly.

3. Final Answer:

   int *ptr; -> Option D

4. Quick Check:

   Pointer declaration = type *name [OK]

Hint: Remember: type *name; declares a pointer [OK]

Common Mistakes:

• Placing * after the variable name
• Using * before the type keyword
• Omitting the * entirely

2. Which of the following is the correct syntax to declare a pointer to a double variable named p?

easy

A. double *p;

B. p double*;

C. *double p;

D. double p*;

Solution

1. Step 1: Recall pointer declaration rules

   The pointer declaration syntax is type *pointerName;. The asterisk (*) must come immediately before the pointer name.

2. Step 2: Evaluate options

   double *p; correctly declares double *p;. Other options have misplaced asterisks or wrong order.

3. Final Answer:

   double *p; -> Option A

4. Quick Check:

   Pointer to double = double *p; [OK]

Hint: Put * right before pointer name, after type [OK]

Common Mistakes:

• Writing * after the variable name
• Placing * before the type keyword
• Swapping variable and type order

3. What is the output of the following code?

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

medium

A. Garbage value

B. Address of x

C. 10

D. Compilation error

Solution

1. Step 1: Understand pointer initialization

   The pointer p is assigned the address of x using &x. So p points to x.

2. Step 2: Dereference pointer to get value

   The expression *p accesses the value stored at the address p points to, which is 10.

3. Final Answer:

   10 -> Option C

4. Quick Check:

   *p gives value at address = 10 [OK]

Hint: Dereference pointer *p to get variable value [OK]

Common Mistakes:

• Printing pointer variable instead of dereferenced value
• Confusing & (address) with * (dereference)
• Expecting address output instead of value

4. Identify the error in the following code snippet:

int a = 5;
int *ptr;
*ptr = a;

medium

A. Pointer ptr is not initialized before dereferencing

B. Variable a is not declared

C. Incorrect pointer declaration syntax

D. Missing semicolon after declaration

Solution

1. Step 1: Check pointer initialization

   The pointer ptr is declared but not assigned any address before dereferencing.

2. Step 2: Understand dereferencing uninitialized pointer

   Dereferencing *ptr without initialization leads to undefined behavior or runtime error.

3. Final Answer:

   Pointer ptr is not initialized before dereferencing -> Option A

4. Quick Check:

   Uninitialized pointer dereference = error [OK]

Hint: Always initialize pointer before using *ptr [OK]

Common Mistakes:

• Dereferencing pointer before assigning address
• Assuming declaration initializes pointer
• Ignoring runtime errors from uninitialized pointers

5. Given the code below, which pointer declaration correctly allows p to point to a constant integer value?

const int val = 100;
// Which declaration is correct?

hard

A. int *p = &val;

B. const int *p = &val;

C. int *const p = &val;

D. int *p const = &val;

Solution

1. Step 1: Understand pointer to const int

   To point to a constant integer, the pointer must be declared as pointing to a const int to prevent modification.

2. Step 2: Analyze options

   int *p = &val; cannot assign const int* to int* (discards qualifiers). const int *p = &val; correctly declares pointer to const int. int *const p = &val; declares const pointer to int (type mismatch). int *p const = &val; invalid syntax.

3. Final Answer:

   const int *p = &val; -> Option B

4. Quick Check:

   Pointer to const int = const int *p [OK]

Hint: Use const int *p to point to constant int [OK]

Common Mistakes:

• Using int *p to point to const int
• Confusing const pointer and pointer to const
• Misplacing const keyword