Write a C Program to implement Linked File Allocation method.

Write a C Program to implement Linked File Allocation method.

Description of Linked allocation:

It is easy to allocate the files because allocation is on an individual block basis. Each block contains a pointer to the next free block in the chain. Here also the file allocation table consisting of a single entry for each file. Using this strategy any free block can be added to a chain very easily. There is a link between one block to another block, that’s why it is said to be linked allocation. We can avoid the external fragmentation.

Algorithm for Linked File Allocation:

Step 1: Create a queue to hold all pages in memory

Step 2: When the page is required replace the page at the head of the queue

Step 3: Now the new page is inserted at the tail of the queue

Step 4: Create a stack

Step 5: When the page fault occurs replace page present at the bottom of the stack

Step 6: Stop the allocation.

/* Program to simulate linked file allocation strategy */

Program Code:

#include<stdio.h>

#include<conio.h>

#include<stdlib.h>

void main()

{

int f[50], p,i, st, len, j, c, k, a;

clrscr();

for(i=0;i<50;i++)

f[i]=0;

printf("Enter how many blocks already allocated: ");

scanf("%d",&p);

printf("Enter blocks already allocated: ");

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

{

scanf("%d",&a);

f[a]=1;

}

x: printf("Enter index starting block and length: ");

scanf("%d%d", &st,&len);

k=len;

if(f[st]==0)

{

for(j=st;j<(st+k);j++)

{

if(f[j]==0)

{

f[j]=1;

printf("%d-------->%d\n",j,f[j]);

}

else

{

printf("%d Block is already allocated \n",j);

k++;

}

}

}

else

printf("%d starting block is already allocated \n",st);

printf("Do you want to enter more file(Yes - 1/No - 0)");

scanf("%d", &c);

if(c==1)

goto x;

else

exit(0);

getch();

}

Program Output:

Enter how many blocks already allocated: 3

Enter blocks already allocated: 1 3 5

Enter index starting block and length: 2 2

2-------->1

3 Block is already allocated

4-------->1

Do you want to enter more file(Yes - 1/No - 0)0