Through the use of examples, we will learn about pointers in C++ and how they operate in this tutorial.

Addresses are symbolically represented by pointers. They give programs the ability to generate and manipulate dynamic data structures in addition to simulating call-by-reference.

Learning C++ pointers is simple and enjoyable. Pointers make some C++ operations easier to complete while preventing others, like dynamic memory allocation, from being completed.

Pointers are variables in C++ that keep track of other variables’ memory addresses.

Address in C++

If we have a variable named var in our program, the command &var will provide its memory address. For instance,

Every variable is a memory location, and each memory location has a determined address that may be accessed using the ampersand (&) operator, which stands for a memory address. Take into account the following, which will print the defined variables’ addresses:

Example 1: Printing Variable Addresses in C++

include

using namespace std;

int main()
{
// declare variables
int var1 = 3;
int var2 = 24;
int var3 = 17;

// print address of var1
cout << "Address of var1: "<< &var1 << endl;

// print address of var2
cout << "Address of var2: " << &var2 << endl;

// print address of var3
cout << "Address of var3: " << &var3 << endl;
}

Output

Address of var1: 0x7fff5fbff8ac
Address of var2: 0x7fff5fbff8a8
Address of var3: 0x7fff5fbff8a4

The hexadecimal form of the address is indicated by the initial 0x in this case.

Keep in mind that the first address is 4 bytes different from the second, and the second address is 4 bytes different from the third.

This is due to the fact that an int variable in a 64-bit system has a size of 4 bytes.

Note: If you run the program, you might not obtain the same outcomes.

C++ Pointers

Pointers are used to hold addresses rather than values, as was already mentioned.

This is how pointers can be declared.

int *pointVar;

Here, an int-type pointer named pointVar has been declared.

The following is another way that pointers can be declared.

int* pointVar; // preferred syntax

Let’s look at another declaration of pointers example.

int* pointVar, p;

Here, a pointer named pointVar and a regular variable named p has been declared.

For the purpose of declaring pointers, the * operator is used after the data type.

Giving Pointers Addresses

Here’s how we can give pointers addresses:

int* pointVar, var;
var = 5;

// assign address of var to pointVar pointer
pointVar = &var;

The variable var in this case has the value 5. Additionally, the code pointVar = &var assigns the address of var to the pointVar pointer.

Utilizing Pointers, obtain the Value from the Address

The * operator is used to retrieve the value pointed to by a pointer. For instance:

int* pointVar, var;
var = 5;

// assign address of var to pointVar
pointVar = &var;

// access value pointed by pointVar
cout << *pointVar << endl;   // Output: 5

The address of var is given to pointVar in the code above. To retrieve the value kept at that address, we used the *pointVar.

The dereference operator is the term used to describe the use of * with pointers. It uses a pointer as its primary mechanism and outputs the value indicated by the address saved in the pointer. In other words, *pointVar = var

Note that pointVar and *pointVar are entirely different in C++. We are unable to use the syntax *pointVar = &var;.

Example 2: Working of C++ Pointers

#include <iostream>
using namespace std;
int main() {
    int var = 5;

    // declare pointer variable
    int* pointVar;

    // store address of var
    pointVar = &var;

    // print value of var
    cout << "var = " << var << endl;

    // print address of var
    cout << "Address of var (&var) = " << &var << endl
         << endl;

    // print pointer pointVar
    cout << "pointVar = " << pointVar << endl;

    // print the content of the address pointVar points to
    cout << "Content of the address pointed to by pointVar (*pointVar) = " << *pointVar << endl;
    
    return 0;
}

Output

var = 5
Address of var (&var) = 0x61ff08
pointVar = 0x61ff08
Content of the address pointed to by pointVar (*pointVar) = 5

Value Pointed by Pointers Changing In C++

If pointVar points to var’s address, we can use *pointVar to modify var’s value.

For instance,

int var = 5;
int* pointVar;

// assign address of var
pointVar = &var;

// change value at address pointVar
*pointVar = 1;

cout << var << endl; // Output: 1

We can modify the value of var by using *pointVar if pointVar points to the address of var.

For illustration

Example 3: Changing Value Pointed by Pointers In C++

#include <iostream>
using namespace std;
int main() {
    int var = 5;
    int* pointVar;

    // store address of var
    pointVar = &var;

    // print var
    cout << "var = " << var << endl;

    // print *pointVar
    cout << "*pointVar = " << *pointVar << endl
         << endl;

    cout << "Changing value of var to 7:" << endl;

    // change value of var to 7
    var = 7;

    // print var
    cout << "var = " << var << endl;

    // print *pointVar
    cout << "*pointVar = " << *pointVar << endl
         << endl;

    cout << "Changing value of *pointVar to 16:" << endl;

    // change value of var to 16
    *pointVar = 16;

    // print var
    cout << "var = " << var << endl;

    // print *pointVar
    cout << "*pointVar = " << *pointVar << endl;
    return 0;
}

Output

var = 5
*pointVar = 5
Changing value of var to 7:
var = 7
*pointVar = 7
Changing value of *pointVar to 16:
var = 16
*pointVar = 16

Common errors made when using pointers In C++

Let’s say we want a pointer variable. Point to point to var’s address Then,

int var, *varPoint;

// Wrong! 
// varPoint is an address but var is not
varPoint = var;

// Wrong!
// &var is an address
// *varPoint is the value stored in &var
*varPoint = &var;

// Correct! 
// varPoint is an address and so is &var
varPoint = &var;

 // Correct!
// both *varPoint and var are values
*varPoint = var;

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