in C language

$8\mathrm{.}14$ LAB: Warm up: Contacts

A linked list is built in this lab. Make sure to keep track of the head node.

Step $1$: Complete main.c to define the struct ContactNode with the following data members:

char contactName$[]$

char contactPhoneNumber$[]$

ContactNode$*$ nextNodePtr

Step $2$: Add the following functions to main.c:

void InitializeContactNode$($ContactNode$*$ thisNode, char$*$ initialName, char$*$ initialPhoneNumber$)-$ Initialize thisNode with initialName and initialPhoneNumber $(1$ tpt$)$

char$*$ GetName$($ContactNode$*$ thisNode$)-$ Return contactName of thisNode $(1$ pt$)$

char$*$ GetPhoneNumber$($ContactNode$*$ thisNode$)-$ Return contactPhoneNumber of thisNode $(1$ pt$)$

void InsertAfter$($ContactNode$*$ thisNode, ContactNode$*$ newContactNode$)-$ Insert newContactNode after thisNode $(2$ pts$)$

ContactNode$*$ GetNext$($ContactNode$*$ thisNode$)-$ Return the next node of thisNode $(1$ pt$)$

void PrintContactNode$($ContactNode$*$ thisNode$)-$ Output contactName and contactPhoneNumber of the thisNode according to the format shown in the example below.

Ex: if the name is Roxanne Hughes and the phone number is $443-555-2864,$ PrintContactNode$()$ outputs:

Name: Roxanne Hughes

Phone number: $443-555-2864$

Step $3$: Define main$()$ to read the name and phone number for three contacts and output each contact. Create three ContactNodes and use the nodes to build a linked list. $(2$ pts$)$

Ex: If the input is:

Roxanne Hughes

$443-555-2864$

Juan Alberto Jr$.$

$410-555-9385$

Rachel Phillips

$310-555-6610$

the output is:

Person $1$: Roxanne Hughes, $443-555-2864$

Person $2$: Juan Alberto Jr$.,410-555-9385$

Person $3$: Rachel Phillips, $310-555-6610$

Step $4$: Output the linked list in main$().$ Use a loop to print each node with PrintContactNode$().(2$ pts$)$

Ex:

CONTACT LIST

Name: Roxanne Hughes

Phone number: $443-555-2864$

Name: Juan Alberto Jr$.$

Phone number: $410-555-9385$

Name: Rachel Phillips

Phone number: $310-555-6610$