Sequential (contiguous) file allocation.Adding on to below C code to print the attached sample program output.#include<stdio.h>#include<stdlib.h>#define TOTAL_DISK_BLOCKS 32#define TOTAL_DISK_INODES 8int blockStatus[TOTAL_DISK_BLOCKS]; // free = 0int blockStart;struct file_table {char fileName[20];int startBlock;int fileSize;int allotStatus;};struct file_table fileTable[TOTAL_DISK_BLOCKS - TOTAL_DISK_INODES];int AllocateBlocks(int Size) {int i = 0, count = 0, inList = 0, nextBlock = 0;int allocStartBlock = TOTAL_DISK_INODES;int allocEndBlock = TOTAL_DISK_BLOCKS - 1;// check whether sufficient free blocks are availablefor (i = 0; i < (TOTAL_DISK_BLOCKS - TOTAL_DISK_INODES); i++)if (blockStatus[i] == 0)count++;if (count < Size)return 1; // not enough free blockscount = 0;while (count < Size) {nextBlock = (rand() % (allocEndBlock - allocStartBlock + 1)) + allocStartBlock;for (i = nextBlock; i < (nextBlock + Size); i++){if (blockStatus[i] == 0)count = count + 1;else {count = 0;break;}}}blockStart = nextBlock;if (count == Size)return 0; // successelsereturn 1; // not successful}void main(){int i =0, j = 0, numFiles = 0, nextBlock = 0, ret = 1;char s[20];---for(i = 0; i < numFiles; i++) {---ret = AllocateBlocks(fileTable[i].fileSize);---}---//Seed the pseudo-random number generator used by rand() with the value seedsrand(1234);---} File allocation method: SEQUENTIALTotal blocks: 32File allocation start at block: 8File allocation end at block: 31Size (kB) of each block: 1Enter no of files: 3Enter the name of file #1: data.csvEnter the size (kB) of file #1: 3Enter the name of file #2: info.docEnter the size (kB) of file #2: 5Enter the name of file #3: music.mp3Enter the size (kB) of file #3: 4FILE_fileNamedata.csvinfo.docmusic.mp3File allocation completed. Exiting.FILE_SIZEBLOCKS_OCCUPIED314-15-16519-20-21-22-23427-28-29-30 File NameFile size (kB)data.csv3info.doc5music.mp34Block Number 0 1 23.4567Sequential allocationFile Name File size (kB) Start blockdata.csvinfo.docmusic.mp3XXBlocks allocatedBlock Number910111213141516 17181914-15-1619-20-21-22-2327-28-29-301419Block Number 2021 22 23 24252627 28 29 30 31427Task:1. Write a C program to simulate the sequential file allocation in a very simple file system.2. Assume a disk of 32 blocks, each block is of 1 KB size3. First 8 blocks (0 to 7) are allocated to the "iNodes" and can't be used by the file system. Hence blocksavailable for allocation are from block 8 to block 31.4. Minimum file size is 1 KB. Hence the file system can have minimum of one file of size 24 KB or maximumof 24 files.5. At the start, it is assumed that the file system has no files.6. The program shall ask the user to input the number of files to allocate and their respective names andfile sizes.7. The program shall randomly (set a seed with srand(seed) to replicate the randomness) select any freeblock as a start block. Check that the start block and the required contiguous blocks are free. If free,allocate those blocks to the file. If not free, find next available contiguous blocks.8. After allocating blocks for all the files, the program shall print file name, file size, and the contiguouslyallocated blocks for each file. Refer to the program output shown below.9. Required test case: Use the example file names and sizes shown above10. Not required test cases:o If there are not enough contiguous blocks available for a file, the program canProgram need not implement file deletion or modification.exit.

Here, I have to provide a C solution to the above question.

Task: 1. Write a C program to simulate the sequential file allocation in a very simple file system. 2. Assume a disk of 32 blocks, each block is of 1 KB size 3. First 8 blocks (0 to 7) are allocated to the "iNodes" and can't be used by the file system. Hence blocks available for allocation are from block 8 to block 31. 4. Minimum file size is 1 KB. Hence the file system can have minimum of one file of size 24 KB or maximum of 24 files. 5. At the start, it is assumed that the file system has no files. 6. The program shall ask the user to input the number of files to allocate and their respective names and file sizes. 7. The program shall randomly (set a seed with srand(seed) to replicate the randomness) select any free block as a start block. Check that the start block and the required contiguous blocks are free. If free, allocate those blocks to the file. If not free, find next available contiguous blocks. 8. After allocating blocks for all the files, the program shall print file name, file size, and the contiguously allocated blocks for each file. Refer to the program output shown below. 9. Required test case: Use the example file names and sizes shown above 10. Not required test cases: o If there are not enough contiguous blocks available for a file, the program can exit. Program need not implement file deletion or modification.

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Sequential (contiguous) file allocation.

Adding on to below C code to print the attached sample program output.

#include<stdio.h>
#include<stdlib.h>
#define TOTAL_DISK_BLOCKS 32
#define TOTAL_DISK_INODES 8
int blockStatus[TOTAL_DISK_BLOCKS]; // free = 0
int blockStart;
struct file_table {
char fileName[20];
int startBlock;
int fileSize;
int allotStatus;
};
struct file_table fileTable[TOTAL_DISK_BLOCKS - TOTAL_DISK_INODES];
int AllocateBlocks(int Size) {
int i = 0, count = 0, inList = 0, nextBlock = 0;
int allocStartBlock = TOTAL_DISK_INODES;
int allocEndBlock = TOTAL_DISK_BLOCKS - 1;
// check whether sufficient free blocks are available
for (i = 0; i < (TOTAL_DISK_BLOCKS - TOTAL_DISK_INODES); i++)
if (blockStatus[i] == 0)
count++;
if (count < Size)
return 1; // not enough free blocks
count = 0;
while (count < Size) {
nextBlock = (rand() % (allocEndBlock - allocStartBlock + 1)) + allocStartBlock;
for (i = nextBlock; i < (nextBlock + Size); i++)
{
if (blockStatus[i] == 0)
count = count + 1;
else {
count = 0;
break;
}
}
}
blockStart = nextBlock;
if (count == Size)
return 0; // success
else
return 1; // not successful
}
void main()
{
int i =0, j = 0, numFiles = 0, nextBlock = 0, ret = 1;
char s[20];
---
for(i = 0; i < numFiles; i++) {
---
ret = AllocateBlocks(fileTable[i].fileSize);
---
}
---
//Seed the pseudo-random number generator used by rand() with the value seed
srand(1234);
---
}

File allocation method: SEQUENTIAL
Total blocks: 32
File allocation start at block: 8
File allocation end at block: 31
Size (kB) of each block: 1
Enter no of files: 3
Enter the name of file #1: data.csv
Enter the size (kB) of file #1: 3
Enter the name of file #2: info.doc
Enter the size (kB) of file #2: 5
Enter the name of file #3: music.mp3
Enter the size (kB) of file #3: 4
FILE_fileName
data.csv
info.doc
music.mp3
File allocation completed. Exiting.
FILE_SIZE
BLOCKS_OCCUPIED
3
14-15-16
5
19-20-21-22-23
4
27-28-29-30

File Name
File size (kB)
data.csv
3
info.doc
5
music.mp3
4
Block Number 0 1 2
3.
4
5
6
7
Sequential allocation
File Name File size (kB) Start block
data.csv
info.doc
music.mp3
X
X
Blocks allocated
Block Number
9
10
11
12
13
14
15
16 17
18
19
14-15-16
19-20-21-22-23
27-28-29-30
14
19
Block Number 20
21 22 23 24
25
26
27 28 29 30 31
4
27
Task:
1. Write a C program to simulate the sequential file allocation in a very simple file system.
2. Assume a disk of 32 blocks, each block is of 1 KB size
3. First 8 blocks (0 to 7) are allocated to the "iNodes" and can't be used by the file system. Hence blocks
available for allocation are from block 8 to block 31.
4. Minimum file size is 1 KB. Hence the file system can have minimum of one file of size 24 KB or maximum
of 24 files.
5. At the start, it is assumed that the file system has no files.
6. The program shall ask the user to input the number of files to allocate and their respective names and
file sizes.
7. The program shall randomly (set a seed with srand(seed) to replicate the randomness) select any free
block as a start block. Check that the start block and the required contiguous blocks are free. If free,
allocate those blocks to the file. If not free, find next available contiguous blocks.
8. After allocating blocks for all the files, the program shall print file name, file size, and the contiguously
allocated blocks for each file. Refer to the program output shown below.
9. Required test case: Use the example file names and sizes shown above
10. Not required test cases:
o If there are not enough contiguous blocks available for a file, the program can
Program need not implement file deletion or modification.
exit.

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