Introduction to Operating Systems with Dr. Ramamurthy Substitution lecture: Project Tips, IPC Scott...
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Transcript of Introduction to Operating Systems with Dr. Ramamurthy Substitution lecture: Project Tips, IPC Scott...
Introduction to Operating Systemswith Dr. Ramamurthy
Substitution lecture: Project Tips, IPCScott Settembre, TASeptember 21, 2010
Project Tips
• What does a process look like?• Pointers and Arrays? Those are easy!• What is a “segmentation fault”?– Two main issues that you will have– Debugging a segmentation fault
Visualize a Process space
Code
Data
Global variables
Stack
Program Counter (PC)
Code
Line 1Line 2Line 3Line 4…
Code
Data
Global variables
Stack
Code
Line 1Line 2Line 3Line 4…
Where are variables stored?#include <stdio.h>
int a_counter = 1;
int main(int argc, char *argv[]) {
int b_counter = 10;
}
What is a pointer in C?int a = 10;…a++;printf(“%i”,a); 11
int * b;b = &a;…*b++;printf(“%i”,a); 12
printf(“%i”,*b);
12
12
b++;printf(“%i”,a);
printf(“%i”,*b); Possible segmentation fault
“a” is an integer
“b” is a pointer to an integer“b” is now pointing to the location in memory that stores “a”
“b” uses the “*” operator to reference the memory it points to
The address that “b” holds, is incremented by the length of an int
What is an array in C?char c = ‘a’;
char ca[10] = “hello”;
a
h e l l o null ? ? ? ?
char * cb;cb = &ca[0]; “cb” is pointing to the first character of the “ca” char array
printf(“%c”,c); a
printf(“%c”,ca[0]); h
printf(“%s”,ca); hello
printf(“%s”,cb); hello
printf(“%s”,&ca[0]); hello
printf(“%c”,&ca[0]); compiler error
Note: zero term
inated string
automatically done for constant
assignment
How do pointers and arrays relate?
• Easy! In C, a pointer IS AN array and an array IS A pointer!
• Differences?– Using the sizeof() operator for an array will give
the total size of the array, but for a pointer it will give the size of a pointer
– The sizeof() operator is a compiler operator and not a callable function
Code issue : Segmentation Faults
• You get a “segmentation fault” when you:– Try to access memory that you are not allowed in– Change a pointer incorrectly– Pass a pointer to a local variable (remember, local
variables are on the stack and subject to disappearing when the go out of scope)
• “Bus error” is like a segmentation fault, but from misuse of the stack
Example of a “Seg Fault”char * arg[2];const char * alphabet = ”abcdefghijklmnopqrstuvwxyz”;strcpy(arg[0],alphabet);
Visualize arg[2]: char * char *
arg[0] arg[1]
4 bytes
Some other memory…
Visualize alphabet: a
1 byte
b c d e x y z null Some other memory………
strc
py Possible segmentation fault
Some other memory…
Indirect example of a “Seg Fault”char * arg[2];int * fa [100]; // Very important financial dataconst char * alphabet = ”abcdefghijklmnopqrstuvwxyz”;strcpy(arg[0],alphabet);
Visualize our memory layout:
int * int *
arg[0] arg[1]
int * int * int * int *
fa[0] fa[1] fa[2] fa[3]
strc
py
printf(“Scott’s 401K plan value: %i”, *fa[0]); Possible segmentation fault
Advice on debugging faults
• For project 1, most faults are due to:– Not using a char array or allocating memory for a
char * to use– Passing a local pointer back from a function– Not zero-terminating your strings– Parsing a string, storing the beginning of a “token”,
but then using that pointer in a string.h function which expects a NULL terminated string
Inter-process Communication
• Process vs. Thread coding issues– Shared memory, shared resources
• What is an “atomic” instruction?• How can I use a semaphore properly?– Protect with a semaphore– Signal with a semaphore– What is going on behind the scenes?
• Example: Larry, Curly, and Moe IPC problem
Visualize Multi-processes
Code
Data
Line 1Line 2Line 3Line 4…
Thread Pool
PC 1
1
Code
Data
Line 1Line 2Line 3Line 4…
Thread Pool
PC 1
1
Atomic instructions// Simple additionint a = 5;a = a + 1; expand this into atomic
instructions 1. Value of “a” from memory into register2. Register gets incremented by 13. Register is put back into memory
Code
Data
a -> rr ++r -> a
a == 5
PC
a == 6
THE Problem that arises
Code
Data
while loop {
a -> rr ++r -> a
}
Thread Pool
PC 1
1 2
PC 2
For example, let’s say:a == 10
There will be a case where:
PC1 runs ‘a->r’ so r==10PC2 runs ‘a->r’ so r==10PC1 runs ‘r++’ so r==11PC2 runs ‘r++’ so r==11PC1 runs ‘r->a’ so a==11PC2 runs ‘r->a’ so a==11
But this is BAD, since two additions occurred!
The value of ‘a’ should be 12!
Concurrency Problem
Semaphore
What is a semaphore?
• A programming object in a multiple process/threaded environment that can:– Restrict access to a common resource– Help synchronize processes/threads
Thread 1
Thread 2
Uses printer buffer
Uses printer buffer
Concurrency Problem
Wait
Wait
Signal
Signal
Blocked
What is a mutex?
• It is a “binary semaphore”– A semaphore with only two states: locked/unlocked– Previous example was a binary semaphore
• Short for “Mutual Exclusion”
• You can always use a binary semaphore in place of a mutex, however, you may want to use a mutex for other reasons
Uses of a semaphore
• Simplistically, you can use a semaphore to achieve two goal perspectives:
– Protect a critical resource/critical section so that only N number of processes/threads can access it at a time
– Signal between N number of processes/threads when it is time for another process/thread can proceed
Use #1 : Protection
Thread 1
Thread 2
Thread 3
Some shared
resource or variable
Semaphore
Wait
WaitBlocked
Blocked
Reads or Writes to variable/resource
Process 1
Process 2
Process 3
Semaphore A
Semaphore B
Semaphore C
Use #2 : Signal
Wait
Wait
Wait
Blocked
Blocked
Blocked
Sign
al
Modify
Signal
Modify
Modify
Signal
Visualize a semaphore
Semaphore
Thread 1
Thread 2
Thread N
…..
Some shared
resource or variable
Current value0 1 2 N…..Wait
Wait
Wait
Blocked
Blocked
Thread modifies the resource or variableSignal
Thread modifies the resource or variableBlocked
Signal
Process “starvation”
• Semaphores do not wake blocked processes in any specific order– In other words, it does not use a FIFO queue– This means, starvation of a process can occur
Visualize a “starvation” scenario
Semaphore
Thread 1
Thread 2
Thread N
…..
Some shared
resource or variable
Wait
Wait
Wait
Blocked
Blocked
Thread modifies the resource or variableSignal
Process “deadlocks”
• Semaphores do not prevent deadlock– They can prevent deadlock, if used cleverly (this
will be discussed more in Lecture – “Dining Philosophers”)
Visualize a “deadlock”
Process 1
Process 2
Some shared
resource or variable
“A”
Some shared
resource or variable
“B”
Wait/Request
Wait/Request
Modify/Assigned
Wait/Request
Wait/Request
Modify/Assigned
Blocked BlockedDeadlock!
Example: Larry, Curley, Moe (LCM) IPC problem
• 3 Farmers named Larry, Curley and Moe
Photo order above is: Curley, Moe, and Larry.
LCM Problem
• There is one shovel• Larry and Moe need to use the shovel to dig• Larry only digs holes• Curley only plants seeds in open holes• Moe only fills holes up after seed planted• Larry can only dig “N” holes ahead of Moe– (Why? Because Larry is attached by chain to Moe.
It is super-comedy-riffic, you see!)
What to understand
• There are three farmers (i.e. three processes)
• Two farmers (i.e. two processes) require the same shovel (i.e. share a resource) to get the job done.
• Farmers can work in parallel, but must also work in sequence– For example: Larry can dig a hole, while Curley plants a
different hole, but they cannot dig and plant the same one.– Also, Curley can plant a hole, while Moe fills an already planted
hole, but they cannot plant and fill the same one.
Visualize LCM
Larry
Curley
Moe
Or we can say, 3 threads running 3 different functions.
Shovel
Or we can say, the shovel is a critical resource (or a critical section).
Think “protect” with a semaphore.
Since only one process can use at a time, think “binary semaphore” or “mutex”.
Holes
Also a critical resource, since for any specific hole, only one farmer can work with it.
However, think “signal” here instead of “protect”. Have the farmers communicate with each other when done with a hole.
Let’s build the code
Larry (digger) Curley (planter) Moe (filler)
Larry digs holewait(Shovel)
signal(Shovel)Moe fills holewait(Shovel)
signal(Shovel)
signal(Curley2go)
wait(Curley2go)
signal(Moe2go)
wait(Moe2go)
Curley plants hole
Hole digging limitation
Larry (digger)
Larry digs holewait(Shovel)
signal(Shovel)
signal(Curley2go)
Let’s pretend Curley and Moe are napping in the field.How many holes would Larry be allowed to dig?
According to the problem, he can only dig N holes ahead of Moe. So if we create a semaphore called “DigHole”, have Larry wait on it, and then signal it N times, he would then dig N holes!
wait(DigHole)
How do we signal the semaphore N times?
We just initialize “DigHole” to have an initial value of N!
The Elegant coding solution
Have 4 semaphores: DigHole = N, Curley2go = 0, Moe2go = 0, and Shovel = 1.
Larry (digger) Curley (planter) Moe (filler)
Larry digs holewait(Shovel)
signal(Shovel)Moe fills holewait(Shovel)
signal(Shovel)
signal(Curley2go)
wait(Curley2go)
signal(Moe2go)
wait(Moe2go)
Curley plants hole
wait(DigHole)
signal(DigHole)