Recitation 8 (Nov. 1)
Outline
TA: Kun Gao
 Lab 5 hints
 Virtual Memory
Reminder
 Shell Lab:
 Due THIS Thursday
Modified from
Minglong Shao’s
Recitation, Fall 2004
Lab 5
 A tiny shell (tsh) with job control & I/O redirection
 Key points:






Understand process tree
Executing programs
Reap all child processes
Handle SIGCHLD, SIGTSTP, SIGINT
Avoid race hazard
I/O redirection
Process tree for shell
pid=10
pgid=10
Shell
Background
job #1
pid=20 Forepgid=20 ground
job
pid=32
pgid=32
Background
process group 32
Child
pid=21
pgid=20
Child
pid=22
pgid=20
Foreground
process group 20
Background
job #2
pid=40
pgid=40
Backgroud
process group 40
Each job has a unique process group id
int setpgid(pid_t pid, pid_t pgid);
setpgid(0, 0);
Process tree for shell (part 2)
pid=5
pgid=5
pid=10
pgid=10
UNIX
shell
tsh
Background
job #1
pid=20 Forepgid=20 ground
job
Foreground job
receives SIGINT, SIGTSTP,
when you type ctrl-c, ctrl-z
pid=32
pgid=32
Background
process group 32
Child
pid=21
pgid=20
Child
pid=22
pgid=20
Foreground
process group 20
Background
job #2
pid=40
pgid=40
Backgroud
process group 40
int kill(pid_t pid, int sig)
pid > 0: send sig to specified process
pid = 0: send sig to all processes in same group
pid < -1: send sig to group -pid
Execute program
int execve(const char *fname,
char *const argv[],
char *const envp[]);
Examples:
execve(“/bin/ls”, NULL, NULL);
execve(“./mytest”, argv, envp);
Reaping child process
waitpid(pid_t pid, int *status, int options)
pid: wait for child process with pid (process id or group id)
-1: wait for any child process
status: tell why child terminated
options:
WNOHANG: return immediately if no children zombied
WUNTRACED: report status of terminated or stopped children
In Lab 5, use
waitpid(-1, &status, WNOHANG|WUNTRACED)
In sigchld_handler():
while ((c_pid = waitpid(…)) > 0) { … }
Reaping child process (part 2)
 int status;
waitpid(pid, &status, WNOHANG|WUNTRACED)
What to check in sigchld_handler:
 WIFEXITED(status): child exited normally (the child
finished, and quit normally on its own)

WEXITSTATUS(status): returns code when child
exits
 WIFSIGNALED(status): child exited because a signal is
not caught (SIGINT, or typing CTRL-C)
 WTERMSIG(status): gives the terminating signal
number
 WIFSTOPPED(status): child is stopped by the receipt of a
signal (SIGSTOP, or typing CTRL-Z)
 WSTOPSIG(status): gives the stop signal number
Reaping child process (part 3)
 Where to put waitpid(…)
 In sigchld_handler(): for both
 tsh should still wait for fg job to complete, how?
Busy wait for foreground job
In eval():
if(fork() != 0) { /* parent */
addjob(…);
while(fg process still alive){
/* do nothing */
}
}
Sleep
In eval():
if(fork() != 0) { /* parent */
addjob(…);
while(fg process still alive){
sleep(1);
}
}
Race hazard
 A data structure is shared by two pieces of code
that can run concurrently
 Different behaviors of program depending upon
how the schedule interleaves the execution of
code.
An example of race hazard
sigchld_handler() {
while ((pid = waitpid(…)) > 0){
deletejob(pid);
}
}
eval() {
pid = fork();
if(pid == 0)
{ /* child */
execve(…);
}
/* parent */
/* signal handler may run BEFORE addjob()*/
addjob(…);
}
An okay schedule
time
Shell
Signal Handler
Child
fork()
addjob()
execve()
exit()
sigchld_handler()
deletejobs()
A problematic schedule
time
Shell
Signal Handler
Child
fork()
execve()
exit()
sigchld_handler()
deletejobs()
addjob()
Job added to job list after the signal handler tried to delete it!
Solution: blocking signals
sigchld_handler() {
pid = waitpid(…);
deletejob(pid);
}
eval() {
sigprocmask(SIG_BLOCK, …)
pid = fork();
if(pid == 0)
{ /* child */
sigprocmask(SIG_UNBLOCK, …)
execve(…);
}
/* parent */
/* signal handler might run BEFORE addjob() */
addjob(…);
sigprocmask(SIG_UNBLOCK, …)
}
Solution: blocking signals (part 2)
 What should our block set be?



SIGSTP
SIGINT
SIGCHLD
I/O redirection
 Do it before call execve() in child process
eval() {
pid = fork();
if(pid == 0)
{ /* child */
/* Redirect I/O */
if (redirect_input)
dup2(…);/* redirect STDIN */
if (redirect_output)
dup2(…);/* redirect STDOUT */
execve(…);
}
addjob(…);
}
dup2(int oldfd, int newfd)
 Covered in Chapter 11
oldfd: old file descriptor
newfd: new file descriptor
 Dup2 makes newfd be a copy of the oldfd (i.e.
operations on newfd is done on oldfd instead)
Some examples:
Get input from in_fd instead of standard input
dup2(in_fd, STDIN_FILENO);
Print output to out_fd instead of standard output
dup2(out_fd, STDOUT_FILENO);
Reminders
 Some important system calls:
 fork(), execve(), waitpid(), sigprocmask(),
setpgid(), kill() …
 Check man pages for details about system calls

man 2 kill
 Check return values of all system calls
 STEP by STEP
 Test your shell by typing commands first
 Each trace worth the same (work on easy ones
first!)
 Start now!
Virtual memory
 One of the most important concepts in CS
 Why use virtual memory (VM)





Use RAM as a cache for disk
Easier memory management
Access protection
Enable “partial swapping”
Share memory efficiently
Virtual memory
Memory
Per Process:
0:
1:
Page Table
Virtual
Addresses
0:
1:
Physical
Addresses
CPU
P-1:
N-1:
 Page
 Page table
 Page hits, Page faults
 Demand Paging
Disk
Conceptual address translation
 Higher bits of address (page number) mapped from
virtual addr. to physical addr.
 Lower bits (page offset) stay the same.
p p–1
n–1
virtual page number (VPN)
page offset
0
virtual address
address translation
m–1
p p–1
physical page number (PPN)
0
page offset
Virtual Memory: up to 2n -1 bytes
Physical Memory: up to 2m -1 bytes
physical address
Address translation through page table
 Translation
 Separate (set of) page table(s) per process
 VPN forms index into page table (points to a page table entry)
page table base register
VPN acts
as
table index
if valid=0
then page
not in memory
virtual address
n–1
p p–1
virtual page number (VPN)
page offset
0
valid access physical page number (PPN)
m–1
p p–1
physical page number (PPN)
page offset
physical address
0
Address translation with TLB
n–1
p p–1
0
virtual page number
page offset
valid
.
virtual address
tag physical page number
.
TLB
.
=
TLB hit
physical address
tag
index
valid tag
byte offset
data
Cache
=
cache hit
data
Integrating VM and cache
 Most caches are physically addressed

Why?
 Perform address translation before cache
lookup, could potentially go to memory =*(
 TLB keeps things manageable
1. VA
CPU
4. PA
MMU/
Translation
2. VPN
3. PTE
TLB
5. Data
Cache/Memory
Integrating VM and cache (TLB hit)
 Most caches are physically addressed

Why?
 Perform address translation before cache
lookup, could potentially go to memory =*(
 TLB keeps things manageable
1. VA
CPU
4. PA
MMU/
Translation
2. VPN
3. PTE
TLB
5. Data
Cache/Memory
Integrating VM and cache (TLB miss)
 TLB miss, but page table is in cache/memory
3. PTEA (PTBP + VPN)
1. VA
CPU
MMU/
Translation
2. VPN
TLB
6. Data
5. PA
4. PTE
Cache/Memory
What is the worst case mem read
(most delay)?
 TLB miss, page fault on page table, then page
fault on memory read
3. PTEA (PTBP + VPN)
1. VA
CPU
MMU/
Translation
2. VPN
5. PA
4. PTE
Cache/Memory
TLB
Page Fault/ Page Fault/
Page Table Mem Read
6. Data
DISK
Example
 Understand end-to-end addr. translation



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20-bit virtual addresses
18-bit physical addresses
Page size is 1024 bytes
TLB is 2-way associative with 16 total entries
Part 1
 A. Virtual address
TLBT
19
18
17
16
TLBI
15
14
13
12
11
10
9
8
7
6
VPN
5
4
3
2
1
0
3
2
1
0
VPO
 B. Physical address
17
16
15
14
13
PPN
12
11
10
9
8
7
6
5
4
PPO
Example: TLB and page table
Part 2
Virtual address 0x04AA4
 A. 04AA4 = 0000 0100 1010 1010 0100
 B. Address translation
parameter
VPN
Value
0x012
Parameter
TLB hit?
Value
Y
TLB Index
TLB Tag
0x2
0x02
Page fault?
PPN
N
0x68
 C. Physical address
01 1010 0010 1010 0100
Part 2
Virtual address 0x78E6
 A. 078E6 = 0000 0111 1000 1110 0110
 B. Address translation
parameter
VPN
Value
0x01E
Parameter
TLB hit?
Value
N
TLB Index
TLB Tag
0x6
0x03
Page fault?
PPN
N
0x57
 C. Physical address
01 0101 1100 1110 0110
End-to-end address translation
 Section 10.6.4: a concrete example
 Read carefully and solve practice problem 10.4