in c$++$

$12\mathrm{.}66$ Chapter $8-$ Shoes w$/$ Pointers

In the given C$++$ code, there are three Shoes objects: shoe$1,$ shoe$2,$ and shoe$3.$

PrintShoes$($shoe$1)$ will print the brands of the shoes in the linked list.

Modify the code to link the shoe$1,$ shoe$2$ and shoe$3$ in a linked list.

using $($and do onot change$)$ the following code:

#include

#include

using namespace std;

$//$ Defining a class named "Shoes"

class Shoes $\{$

public:

string brand; $//$ Member variable to store the brand of the shoe

Shoes$*$ nextShoesAddress; $//$ Pointer to the next Shoes object in the list

$\}$;

$//$ Function to print the list of shoes

void PrintShoes$($Shoes startShoe$)\{$

Shoes$*$ currentShoe; $//$ Pointer to keep track of the current shoe

int shoeCount; $//$ Variable to keep count of the shoes

currentShoe $=$ &startShoe; $//$ Set the current shoe to the start of the list

shoeCount $=0$;

$//$ Loop through the list of shoes until we reach the end

while $($currentShoe $!=0)\{$

shoeCount $=$ shoeCount $+1$;

cout $<<$ "Shoe $"<<$ shoeCount $<<"="<<(*$currentShoe$).$brand $<<$ endl;

currentShoe $=(*$currentShoe$).$nextShoesAddress; $//$ Move to the next shoe in the list

$\}$

return;

$\}$

int main$()\{$

Shoes shoe$1,$ shoe$2,$ shoe$3$;

$//$ Assigning brands to each shoe

shoe$1.$brand $=$ "Nike";

shoe$2.$brand $=$ "Puma";

shoe$3.$brand $=$ "Adidas";

$///$ Code here to link shoe$1,$ shoe$2$ and shoe$3$ in a linked list

$///$

$//$ Calling the function to print the list of shoes, starting from shoe$1$

PrintShoes$($shoe$1)$;

return $0$;

$\}$