Question

* C PROGRAMMING*

Outcomes:

• Demonstrate the ability to create and use linked lists in dynamic memory
• Demonstrate the ability to add nodes and remove nodes from a linked list of structs

Program Specifications:

• Write a program that creates the following struct (you can give it whatever name you want):
  • char name[100];
  • int age;
  • float weight;

Create the following menu system:

1. Add a Record
2. Display All Records
3. Quit

When the user selects (1) you will prompt them for a name, age, and weight. You will insert the record into the linked list which will be alphabetically ordered by name (first letter of full name).

When the user selects (2) you will display all records currently in the list.

This program will not use an array, but it will use a linked list of structs. The program will allow for an unlimited number of records. The only limit will the amount of available RAM.

DO NOT USE:

• Use global variables
• Use the goto statement
• Use the break command outside a case statement

Answer 1

Note: The language used is C.

Code:

#include<stdio.h>
#include<string.h>
#include<malloc.h>

struct record{
char name[100];
int age;
float weight;
};

//Linked List Node
struct Node{
struct record rec;
struct Node *next;
};

//This function adds a record in the linked list
//according to alphabetical order of name.
void addRecord(struct Node **head, struct record newRecord){
//Create a new node
struct Node *newNode = (struct Node*)malloc(sizeof(struct Node));
newNode->rec = newRecord;

//When linked list is empty
if(*head == NULL){
newNode->next = NULL;
*head = newNode; //Set head pointer to newNode.
return;
}

//If linked list has only one node.
if((*head)->next == NULL){
//Compare names of newNode and head.
if(strcmp((*head)->rec.name, newRecord.name) > 0){
newNode->next = *head;
*head = newNode;
return;
}
(*head)->next = newNode;
newNode->next = NULL;
}

//Maintain current and previous pointers and traverse the linked list.
struct Node *curr = (*head)->next;
struct Node *prev = *head;
while(curr){
if(strcmp(curr->rec.name, newRecord.name) > 0){
newNode->next = curr;
prev->next = newNode;
return;
}
prev = curr;
curr = curr->next;
}

//If newNode is added to the last.
prev->next = newNode;
newNode->next = NULL;
}

//This function displays all the records.
void displayRecords(struct Node *head){
int i = 1;
struct Node *curr = head;

//Traverse the linked list and print the data.
while(curr){
printf("%d.\n", i);
printf("Name: %s\n", curr->rec.name);
printf("Age: %d\n", curr->rec.age);
printf("Weight: %.2f\n\n", curr->rec.weight);
i++;
curr = curr->next;
}
}

//Main function
int main(){
//head node is initialized to null.
struct Node *head = NULL;
int option = 1;
printf("Select appropriate option.\n");

while(option != 3){
struct record newRecord;
//Menu
printf("1. Add a record\n");
printf("2. Display all records\n");
printf("3. Quit\n");

scanf("%d", &option);
switch(option){
case 1:
printf("Enter name(beginning with capital letter and rest small):\n");
scanf("%s", newRecord.name);
printf("Enter age:\n");
scanf("%d", &newRecord.age);
printf("Enter weight:\n");
scanf("%f", &newRecord.weight);
addRecord(&head, newRecord); //We pass the address of head so that it can be modified by the function.
break;
case 2:
displayRecords(head);
break;
case 3:
break;
default:
printf("Select correct option\n");
break;
}
}
return 0;
}

Code Snippets:

Sample Output: