Question

The topic of this laboratory exercise a probably one of the most ieceresting ones you willsee in thi·coure. The idea of linking together dynamicaby-located memory block, ae.. dynamic vaniables) into a thain of them, a owerlul coneept, athough not an easy one to wnderstand nor to program. So, buckke your veat belts and strap on your belmets cause we re eoing for a hard ride herE, We bgn and doing it at run tise is a very p You now know how to dynamicaly creace blocks of memory at run time and access them through pointers (the only way to do sel Remember that such blocks of memory are really instantations type. It is juat another way to instantiate variables from the struct tomplate, ecept now we do R dynamicaly, Therefore, it i bet to define the structure that we want this variable to look e, a a new data type ving the typedef operation. From this point on, this is the only acceptable way to delne a variable type that be instantiated dynamicaly, so get uned to it of structures of the struct Knowing how to dynamically alocate a block of memory as an instance of a struct template and asigning values to its members is rather strht-torward. You use two eew functions you just leamed (sizeof and llocto instantiate a dynamic variable, a memben uing the arrow operator, and voilal you have a stnuctured vaniable instance with assign values to its variable itstances you w have created so that you can keep them together in a sensible way, and be able to access each of theve blocks individualiy latce for reading for writing or for delkting One way to do that of course. is to put chem in an array However, that defeuts the whole purpose ef Synamicaly allocating variables at tun time, wihout having to inew how large an areay you will need. So, that is not a practical solution The better way to do it a to link the trusture. There are several variatiom of inked struatures teg ete, but the one we will treat here is the knked &s instances together wing pointers and form a nked linked stackx, linked queUe To visualize a linked ist,ehink of elephants in a cireus where one elephant grabs the tail of the one in front of it with its trunk, and its own tail o in turn grabbed by the elephant o on. Thee elephants are Iinked together, tal to trunk to talinto one powby long chain Same for a linked ist. One block (instance veriablel points to the one behind it, which int points to the one after it, and so on. There is one pointer poiming to the first one caled the heod pointer, and the "tai of the lest block in the linked lst points to NULL to indicate the end ofthe kt,Graphica y, 리ooklke this: behind it and

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Answer 1

#include<stdio.h>

struct node

{

int data;

struct node *next;

};

struct node *insert_end(struct node *);

struct node *delete(struct node *);

void display(struct node *);

int main()

{

struct node *start=NULL;

struct node *tmp;

int i;

for(i=0;i<5;i++)

{

tmp=(struct node *)malloc(sizeof(struct node));

tmp->data=4-i;

tmp->next=start;

start=tmp;

}

printf("current list:");

display(start);

start=insert_end(start);

printf("\nnode after insertion at end:");

display(start);

start=delete(start);

}

struct node *insert_end(struct node *start) //inserts node at end of the lis

{

int data;

printf("insert data(node) to be inserted at end");

scanf("%d",&data);

struct node *tmp;

tmp=(struct node *)malloc(sizeof(struct node));

tmp->data=data;

tmp->next=NULL;

struct node *p=start;

if(p==NULL)

{

p->next=tmp;

return p;

}

while(p->next!=NULL)

p=p->next;

p->next=tmp;

return start;

}

struct node *delete(struct node *start) //deletes the desired node

{

int x;

printf("insert the node to be deleted");

scanf("%d",&x);

struct node *tmp=start, *prev;

if(start->data==x)

{

start=start->next;

free(tmp);

return start;

}

while(tmp!=NULL && tmp->data!=x)

{

prev=tmp;

tmp=tmp->next;

}

if(tmp==NULL)

{

printf("node to be deleted is not present in the list\n");

return start;

}

prev->next=tmp->next;

free(tmp);

printf("\nnode after deletion:");

display(start);

return start;

}

void display(struct node *start)

{

struct node *p;

if(start==NULL)

{

printf("Empty\n");

return;

}

p=start;

while(p!=NULL)

{

printf("%d->",p->data);

p=p->next;

}

printf("\n\n");

}