10/24/2010
COP4600 Discussion 8
Segmentation, File System, Fork() Implementation
Reminder

Homework3 is posted on Sakai, it’s due
one week. Please prepare early.
TA: Huafeng Jin
hj0@cise.ufl.edu
Discussion 7 Recap

Assignment 3
 C Version
 Description (IPC producer-consumer)
 IPC in C
(shm: shmget, shmat, shmctl, shmdt)
 Semaphore in C (sem_init, sem_wait, sem_post)
Question-1
 What does the following programs do?
Program1:
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <stdio.h>
main() {
char c, *shm, *s;
int shmid;
key_t key = 5678;
shmid = shmget(key, 27, IPC_CREAT | 0666);
shm = shmat(shmid, NULL, 0);
s = shm;
 Java Version
 Description (simulate 3 algorithms, calculate page faults)
 Page Replacement Algorithms
 Second Chance (R bit, load time, linkedlist)
 NRU
(R bit, M bit, 4 categories)
 Aging
(R bit, counter, right shift and insert R)
for (c = 'a'; c <= 'z'; c++)
*s++ = c;
*s = NULL;
}
Question-1
Question-1(cont)
 What does the following programs do?
Program 2:
Program1:
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <stdio.h>
main() {
int shmid;
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <stdio.h>
main() {
char c, *shm, *s;
key_t key = 5678;
int shmid;
key_t key = 5678;
shmid = shmget(key, 27, IPC_CREAT | 0666);
shm = shmat(shmid, NULL, 0);
s = shm;
char *shm, *s;
shmid = shmget(key, 27, 0666);
shm = shmat(shmid, NULL, 0);
for (s = shm; *s != NULL; s++)
for (c = 'a'; c <= 'z'; c++)
*s++ = c;
putchar(*s);
putchar('\n');
}
*s = NULL;
}
Put 26 letters into a shared memory with key 5678
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Question-1(cont)
Program 2:
Question-2

At the following state, using NRU, what happens?
#include <sys/types.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <stdio.h>
main() {
int shmid;
1) Write 5:
key_t key = 5678;
char *shm, *s;
shmid = shmget(key, 27, 0666);
shm = shmat(shmid, NULL, 0);
for (s = shm; *s != NULL; s++)
putchar(*s);
putchar('\n');
2) Read 6:
}
Get characters from shared memory with key 5678
Question-2

At the following state, using NRU, what happens?
Outline

Segmentation
 Segmentation vs. paging
 Segmentation with paging

1) Write 5:
File System
 File System Implementation
 FAT
 I-node
 File Directory
2) Read 6:
 Implementation with I-node

Outline

Segmentation
 Segmentation vs. paging
 Segmentation with paging

File System
Fork() Implementation
Segmentation


Initially, the memory has only one address space
Segmentation:
◦ Provide the machine with many logically completely
independent address spaces.
 File System Implementation
 FAT
 I-node
 File Directory
 Implementation with I-node

Fork() Implementation
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Segmentation vs. Paging


Segmentation with paging
Segmentation:

Virtual memory:
◦ Programmer is aware of segmentation.
◦ Segmentation can be expanded or shrunk.

Segment:
◦ Comprised of segments
◦ Comprised of pages
Paging:
◦ Programmer is NOT aware of paging.
◦ Page size is fixed.
For other differences, refer to Figure 3-33 (P238)
Segmentation with paging

Virtual address format:
Example

◦ Page size is 4KB
◦ Segment no. + page no. + offset
Example

In virtual space, we have:
◦ Page size is 4KB
◦ 32 pages
◦ Each segment has at most 8 pages
How many bits are needed for segment number in the address?
How many bits are needed for page number in the address?
What is the MULTICS address for 23200?
 Totally 32/8 = 4 segments, so 2 bits for segment no.
 Each segment has 8 pages, so 3 bits for page no.
 23200%4096 = 2720
In virtual space, we have:
◦ 32 pages
◦ Each segment has at most 8 pages
How many bits are needed for segment number in the address?
How many bits are needed for page number in the address?
What is the MULTICS address for 23200?
Outline

Segmentation
 Segmentation vs. paging
 Segmentation with paging

File System
 File System Implementation
 FAT
 I-node
 File Directory
 Implementation with I-node
 Seg #: 0, Page #: 5, Offset: 2720

Fork() Implementation
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File System

File
File System Implementation

◦ A block of resource for storing information based on
some durable storage.


Methods:
◦ Contiguous Allocation
◦ Linked List Allocation
The basic unit for a file is called block.
File operations:
 FAT (File Allocation Table)
◦ I-nodes
◦ create
◦ delete
◦ open
◦ close
◦ read
◦ write
◦ append
FAT

FAT (File Allocation Table)
◦ Linked list allocation using a table in memory.
 Example:
FAT
Using FAT, for each block, we have to provide an
address.
 Example:

◦ Disk is 64GB
◦ Block size is 16KB
What is the memory size to store file allocation table?
 File A: 4, 7, 2, 10, 12 blocks
 File B: 6, 3, 11, 14 blocks
FAT
Using FAT, for each block, we have to provide an
address.
 Example:

◦ Disk is 64GB
◦ Block size is 16KB
What is the memory size to store file allocation table?
 Number of blocks: 64GB/16KB = 236/214 = 222
So the each address takes 22 bits.
Memory size: 22 * 222 = 92,274,688 bits
= 11,534,336 bytes = 11MB
I-node

I-node:

Remember file management system calls:
◦ List attributes and disk addresses of the file’s blocks.
◦ File descriptor table
◦ File table
◦ I-node table
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Direct and Indirect fields:

Direct fields:

Single indirect fields:
Direct and Indirect fields:
 Records the address of the block
 Records the address of the block that contains the
addresses of blocks.

Double indirect fields:
 Records the address of the block that contains the
addresses of blocks. Each such block contains
address of blocks.

Triple indirect fields: …
Example

Each I-node can store:
Example

◦ 12 direct addresses
◦ 2 single indirect addresses
◦ 1 double indirect address
Block size is: 2KB, block address takes 16bits.
What is the max file size?
Each I-node can store:
◦ 12 direct addresses
◦ 2 single indirect addresses
◦ 1 double indirect address
Block size is: 2KB, block address takes 16bits.
What is the max file size?
 2KB/16bits = 2 11/2 = 2 10
 Each block can hold 210 block addresses.
 Direct address: 12 blocks
 Single indirect address: 2*210 blocks
 Double indirect address: (2 10)2 blocks
Example (cont)

So totally we can hold:
◦ 12 + 2*210 + (210)2 = 1,050,636 blocks
◦ 1,050,636* 2KB ≈ 2.1GB
The maximum file size is 2.1GB
 Remember:
File Directory

Directory:
◦ A virtual container of files

A file hierarchy:

◦ If there is n0 direct address, 1 single indirect, 1 double
indirect, 1 triple indirect, and
◦ Assuming N block addresses can fit in a block, then
 n0 + N + N2 + N3 blocks can be hold at most.
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Implement Directories

With I-nodes:
Implement Directories

◦ Each directory has a directory block
◦ Directory block: (just like file block)
Structure with directory blocks:
◦ Access file f2:
Each entry contains:
 Directory/file name
 Pointer to the i-node
Example

For the following file hierarchy:
Example

Example
Which blocks do we need to read in order to
open f2?
Outline

 Segmentation vs. paging
 Segmentation with paging
◦ 100, 200, 101, 201, 107

If we want to add a new directory d6 under d1,
which blocks need to be updated?

File System
 File System Implementation
◦ 201

Segmentation
 FAT
 I-node
If we want to add 1 more block to f6, which
block need to be updated?
 File Directory
 Implementation with I-node
◦ 206

Fork() Implementation
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10/24/2010
Fork() Implementation

Fork():
◦ Creates a new process.
◦ Makes a copy of the current process’s address space
to the new process.
Fork() Implementation

Fork():
◦ Creates a new process.
◦ Makes a copy of the current process’s address space
to the new process.
Very time consuming, is
there a way to delay the
copy of the address
space?
Recap

Segmentation
 Segmentation vs. paging
 Segmentation with paging

File System
Any Questions?
 File System Implementation
 FAT
 I-node
 File Directory
 Implementation with I-node

Fork() Implementation
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