C$++!$ I need you help to Write down the missing code and test it with attached cases?

using namespace std;

#include

#include"llist.h$"$

llist::llist$()$

$\{$ cout in llist constructor..." $$ endl; $\}$

llist::~llist$()$

$\{$ cout in llist destructor..." $$ endl; $\}$

$//$PURPOSE:

$//$PARAMETER:

bool llist::isEmpty$()$

$\{\}//$ be sure to check all $3$ data members

$//$PURPOSE:

$//$PARAMETER:

void llist::displayAll$()$

$\{\}//$ be sure to display horizontally in $[]$ with

$//$ blanks between elements e$.$g$.[1234]$

$//$ You MUST use while $($P $!=$ NULL$)$ loop or you will not get the credit!

$//$PURPOSE:

$//$PARAMETER:

void llist::addRear$($el$\_$t NewNum$)$

$\{\}//$ comment the $2$ cases

$//$PURPOSE:

$//$PARAMETER:

void llist::addFront$($el$\_$t NewNum$)$

$\{\}//$ comment the $2$ cases

$//$PURPOSE:

$//$PARAMETER:

void llist::deleteFront$($el$\_$t& OldNum$)$

$\{\}//$ comment the $3$ cases

$//$PURPOSE:

$//$PARAMETER:

void llist::deleteRear$($el$\_$t& OldNum$)$

$\{\}//$ comment the $3$ cases

$//$ Utility Function to move a local pointer to the Jth node

void llist::moveTo$($int J$,$ Node$*$& temp$)$

$\{//$ Note that case$1$ client does not test this. But it is neede

$//$ for case $2$ so make it work perfectly.

$//$ moves temp J$-1$ times to go to the Jth node $(>=1$ and Count is assumed$)$

$//$ for $($ int K $=...)$ temp $=$ temp$->$Next;

$\}$

$/*---$ harder ones for case $2$ and $3$ follow $--*/$

$//$PURPOSE:

$//$PARAMETER:

void llist::deleteIth$($int I, el$\_$t& OldNum$)$

$\{\}//$ must use moveTo to move local pointers. Be sure to comment to which node you are moving them. Do not forget to set OldNum.

$//$PURPOSE:

$//$PARAMETER:

void llist::insertIth$($int I, el$\_$t newNum$)$

$\{\}//$ must use moveTo to move local pointers. Be sure to comment to which node you are moving them.

$//$PURPOSE:

$//$PARAMETER:

llist::llist$($const llist& Original$)$

$\{\}//$ use my code below

$//$PURPOSE: This is the copy constructor which copies a list into another.

$//$PARAMETER: Original $-$ it the variable used to pass the list to be copied.

llist::llist$($const llist& Original$)$

$\{//$ use my code

Front $=$ NULL; Rear $=$ NULL; Count $=0$;

$//$ this$->$ object has to be built up by allocating new cells withOtherOne elements $(**)$

Node$*$ P; $//$ local pointer for OtherOne

P $=$ Original.Front;

while $($P $!=$ NULL$)//$ a loop which repeats until you reach the end of OtherOne.

$\{$

this$->$addRear$($P$->$Elem$)$; $//$Ps element is added to this$->$

P $=$ P$->$Next; $//$ Go to the next node in OtherOne

$\}$

return $*$this; $//$ return the result unconditionally. Note that the result is returned by reference.

$\}//$ end of if

$//$PURPOSE:

$//$PARAMETER:

llist& llist::operator$=($const llist& OtherOne$)$

$\{$

el$\_$t x;

$//$ First make sure this$->$ and OtherOne are not the same object.

$//$ To do this, compare the pointers to the objects $.$

if $($&OtherOne $!=$ this$)//$ if not the same

$\{$

$//$ this$->$ object has to be emptied first.

while $(!$ this$->$isEmpty$())$

this$->$deleteRear$($x$)$;

$//$ this$->$ object has to be built up by allocating new cells with OtherOne elements $(**)$

Node$*$ P; $//$ local pointer for OtherOne

P $=$ OtherOne.Front;

while $($P $!=$ NULL$)//$ a loop which repeats until you reach the end of OtherOne.

$\{$

this$->$addRear$($P$->$elem$)$; $//$P$$s element is added to this$->$

P $=$ P$->$Next; $//$ Go to the next node in OtherOne

$\}$

$\}//$ end of if

return $*$this;

$\}//$ use my code above

$1.$check empty and report the result

$2.$display the list

L$.$displayAll$()$;

$3.$add $4$ integers to rear

L$.$addRear$(1)$; L$.$addRear$(2)$; L$.$addRear$(3)$; L$.$addRear$(4$

$4.$display the list

L$.$displayAll$()$;

$-\left[\begin{array}{cccc}1& 2& 3& 4\end{array}\right]$

$5.$remove from front twice $($and display the elements as they are removed$)$

$6.$display the list

$7.$check empty again and report the result

$8.$remove from the rear twice $($display the elements removed$)$

$9.$check empty again and report the result

$10.$add to front $4$ times $($elements $9,8,6,5)$

$11.$display the list

$12.$remove from the rear $4$ times $($display the elements removed$)$

$13.$display the list again