Question

Programing C Just with #include <stdio.h> We will create a singly linked list of 7 nodes. Then, the user will tell us whether to print the “odd-placed” nodes or the “even-placed” nodes. For example, if I had a list of 5 nodes like below, and the user specifies for the odd nodes to be printed, my program would print the values in nodes nl and n3. If the user instead indicated for the even nodes to be printed, my program would print the values in no, n2 and n4. no n1 n2 n3 n4 29 32 81 13 10 Note: a singly-linked-list structure looks something like this (please use different names than the ones provided): struct node{ int value; struct node *after; . Guidelines: Don't worry about functions, you can implement everything in main. Ask the user for the direction (a character). Don't worry about asking for the values of the nodes, you can set them up manually. Use a while loop to print the values, like we did in class. The key is the “after” pointer. How many times should you call this pointer per loop? Also, if you are not careful, the even case might give you a segmentation fault. O . These are some output samples: Odd or even printing (o/e)? O 32 13 Odd or even printing (o/e)? e 29 81 10

Answer 1

C code:

#include<stdio.h>
#include<conio.h>
/* A Linked list node */
struct Node {
int value;
struct Node* after;
};
// function to insert a node at the
// beginning of the linked list
void push(struct Node** head_ref, int new_value)
{
/* allocate node */
struct Node* new_node = (struct Node *)malloc(sizeof(struct Node));

/* put in the value */
new_node->value = new_value;

/* link the old list to the new node */
new_node->after = (*head_ref);

/* move the head to point to the new node */
(*head_ref) = new_node;
}
int main()
{
struct Node* head = NULL;

// create linked list 10->13->81->32->29
//passing reference of head so that changes
//made in push function are reflected in main as well
push(&head, 29);
push(&head, 32);
push(&head, 81);
push(&head, 13);
push(&head, 10);

char option;
printf("Odd or Even printing (o/e)? ");
scanf("%c",&option);
//if option is o then we'll print odd nodes
if(option=='o'){
if(head!=NULL){//checking if there are more than single node so that we can start from second node.
head= head->after;
}else{
return;
}
while(head!=NULL&& head->after!= NULL){
printf("%d\t",head->value);
head= head->after->after;
}
}else if(option=='e'){
//handling case of single node in the linked list
if(head!=NULL&& head->after==NULL){
printf("%d\t",head->value);
}else{
while(head!=NULL&& head->after!=NULL){
printf("%d\t",head->value);
head= head->after->after;
}
}
// handling the last remaining element
if(head!=NULL){
printf("%d\t",head->value);
}
}
return 0;
}

Output:

C:\Users\HP\Desktop\CHEGG\linked List.exe Odd or even printing (o/e)? e 10 81 29 Process returned o (exo) execution time : 1.787 s Press any key to continue.

C:\Users\HP\Desktop\CHEGG\linked List.exe Odd or Even printing (o/e)? O 13 32 Process returned © (@xo) execution time : 4.376 s Press any key to continue.

Note:- \t in the print statement is used for tab type space while printing.

Screenshots of the code:-

X linked List.c X #include<stdio.h> #include<conio.h> /* A Linked list node */ struct Node { int value; struct Node* after; // function to insert a node at the // beginning of the linked list void push(struct Node** head_ref, int new_value) { /* allocate node */ struct Node* new_node = (struct Node *) malloc(sizeof(struct Node)); /* put in the value */ new_node->value = new_value; /* link the old list to the new node */ new_node->after = (*head_ref); /* move the head to point to the new node */ (*head_ref) new_node; int main() { struct Node* head = NULL; // create linked list 10->13->81->32->29 //passing reference of head so that changes //made in push function are reflected in main as well push (Ghead, 29);

Start here X linkedList.c X push (Ghead, 29); push (Ghead, 32); push(Ghead, 81); push(Ghead, 13); push(Ghead, 10); 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 - char option; printf("Odd or Even printing (o/e)? "); scanf("%c", Goption); //if option is o then we'll print odd nodes if (option==''){ if (head!=NULL) { // checking if there are more than single node so that we can start from second node head= head->after; Jelse return; .... Y 48 while(head !=NULLGE head->after!= NULL) { printf("%d\t", head->value); head= head->after-after; 49 50 51 52 53 54 55 0 }else if (option=='e') { //handling case of single node in the linked list if(head !=NULL&& head->after==NULL) { printf("%d\t", head->value); else { while (head !=NULL&S head->after!=NULL) { printf("%d\t", head->value); head= head-after-after, ...... 00 Jun 5 } 57 58 59 60 61 62 63 64 65 0 // handling the last remaining element if (head!=NULL) { printf("%d\t", head->value); ... тат

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