Exercise

Enhance your understanding by getting hands-on practice ​with this exercise on locks!

We'll cover the following

Simulator

This program, x86.py, allows you to see how different thread interleavings either cause or avoid race conditions. See the README for details on how the program works and answer the questions below. You can also recap the details of this simulator from an earlier chapter here.


Welcome to this simulator. The idea is to gain familiarity with threads by
seeing how they interleave; the simulator, x86.py, will help you in
gaining this understanding.

The simulator mimicks the execution of short assembly sequences by multiple
threads. Note that the OS code that would run (for example, to perform a
context switch) is *not* shown; thus, all you see is the interleaving of the
user code.

The assembly code that is run is based on x86, but somewhat simplified. 
In this instruction set, there are four general-purpose registers 
(%ax, %bx, %cx, %dx), a program counter (PC), and a small set of instructions
which will be enough for our purposes. We've also added a few extra GP
registers (%ex, %fx) which don't quite match anything in x86 land
(but that is OK).

Here is an example code snippet that we will be able to run:

.main
mov 2000, %ax   # get the value at the address
add $1, %ax     # increment it
mov %ax, 2000   # store it back
halt

The code is easy to understand. The first instruction, an x86 "mov", simply
loads a value from the address specified by 2000 into the register %ax.
Addresses, in this subset of x86, can take some of the following forms:

  2000          -> the number (2000) is the address
  (%cx)         -> contents of register (in parentheses) forms the address
  1000(%dx)     -> the number + contents of the register form the address
  10(%ax,%bx)   -> the number + reg1 + reg2 forms the address
  10(%ax,%bx,4) -> the number + reg1 + (reg2*scaling) forms the address

To store a value, the same "mov" instruction is used, but this time with the
arguments reversed, e.g.:

  mov %ax, 2000

The "add" instruction, from the sequence above, should be clear: it adds an
immediate value (specified by $1) to the register specified in the second
argument (i.e., %ax = %ax + 1).

Thus, we now can understand the code sequence above: it loads the value at
address 2000, adds 1 to it, and then stores the value back into address 2000.

The fake-ish "halt" instruction just stops running this thread.

Let's run the simulator and see how this all works! Assume the above code
sequence is in the file "simple-race.s".

prompt> ./x86.py -p simple-race.s -t 1 

       Thread 0
1000 mov 2000, %ax
1001 add $1, %ax
1002 mov %ax, 2000
1003 halt

prompt> 

The arguments used here specify the program (-p), the number of threads (-t
1), and the interrupt interval, which is how often a scheduler will be woken
and run to switch to a different task. Because there is only one thread in
this example, this interval does not matter.

The output is easy to read: the simulator prints the program counter (here
shown from 1000 to 1003) and the instruction that gets executed. Note that we
assume (unrealistically) that all instructions just take up a single byte in
memory; in x86, instructions are variable-sized and would take up from one to
a small number of bytes. 

We can use more detailed tracing to get a better sense of how machine state
changes during the execution:

prompt> ./x86.py -p simple-race.s -t 1 -M 2000 -R ax,bx

 2000      ax    bx          Thread 0
    ?       ?     ?
    ?       ?     ?   1000 mov 2000, %ax
    ?       ?     ?   1001 add $1, %ax
    ?       ?     ?   1002 mov %ax, 2000
    ?       ?     ?   1003 halt

Oops! Forgot the -c flag (which actually computes the answers for you).

prompt> ./x86.py -p simple-race.s -t 1 -M 2000 -R ax,bx -c

 2000      ax    bx          Thread 0
    0       0     0
    0       0     0   1000 mov 2000, %ax
    0       1     0   1001 add $1, %ax
    1       1     0   1002 mov %ax, 2000
    1       1     0   1003 halt

By using the -M flag, we can trace memory locations (a comma-separated list
lets you trace more than one, e.g., 2000,3000); by using the -R flag we can
track the values inside specific registers.

The values on the left show the memory/register contents AFTER the instruction
on the right has executed. For example, after the "add" instruction, you can
see that %ax has been incremented to the value 1; after the second "mov"
instruction (at PC=1002), you can see that the memory contents at 2000 are
now also incremented.

There are a few more instructions you'll need to know, so let's get to them
now. Here is a code snippet of a loop:

.main
.top
sub  $1,%dx
test $0,%dx     
jgte .top         
halt

A few things have been introduced here. First is the "test" instruction.
This instruction takes two arguments and compares them; it then sets implicit
"condition codes" (kind of like 1-bit registers) which subsequent instructions
can act upon.

In this case, the other new instruction is the "jump" instruction (in this
case, "jgte" which stands for "jump if greater than or equal to"). This
instruction jumps if the second value is greater than or equal to the first
in the test.

One last point: to really make this code work, dx must be initialized to 1 or
greater. 

Thus, we run the program like this:

prompt> ./x86.py -p loop.s -t 1 -a dx=3 -R dx -C -c

   dx   >= >  <= <  != ==        Thread 0
    3   0  0  0  0  0  0
    2   0  0  0  0  0  0  1000 sub  $1,%dx
    2   1  1  0  0  1  0  1001 test $0,%dx
    2   1  1  0  0  1  0  1002 jgte .top
    1   1  1  0  0  1  0  1000 sub  $1,%dx
    1   1  1  0  0  1  0  1001 test $0,%dx
    1   1  1  0  0  1  0  1002 jgte .top
    0   1  1  0  0  1  0  1000 sub  $1,%dx
    0   1  0  1  0  0  1  1001 test $0,%dx
    0   1  0  1  0  0  1  1002 jgte .top
    0   1  0  1  0  0  1  1003 halt

The "-R dx" flag traces the value of %dx; the "-C" flag traces the values of
the condition codes that get set by a test instruction. Finally, the "-a dx=3"
flag sets the %dx register to the value 3 to start with. 

As you can see from the trace, the "sub" instruction slowly lowers the value
of %dx. The first few times "test" is called, only the ">=", ">", and "!="
conditions get set. However, the last "test" in the trace finds %dx and 0 to
be equal, and thus the subsequent jump does NOT take place, and the program
finally halts.

Now, finally, we get to a more interesting case, i.e., a race condition with
multiple threads. Let's look at the code first:

.main
.top
# critical section
mov 2000, %ax       # get the value at the address
add $1, %ax         # increment it
mov %ax, 2000       # store it back

# see if we're still looping
sub  $1, %bx
test $0, %bx
jgt .top

halt

The code has a critical section which loads the value of a variable 
(at address 2000), then adds 1 to the value, then stores it back. 

The code after just decrements a loop counter (in %bx), tests if it
is greater than or equal to zero, and if so, jumps back to the top
to the critical section again.

prompt> ./x86.py -p looping-race-nolock.s -t 2 -a bx=1 -M 2000 -c

 2000      bx          Thread 0                Thread 1
    0       1
    0       1   1000 mov 2000, %ax
    0       1   1001 add $1, %ax
    1       1   1002 mov %ax, 2000
    1       0   1003 sub  $1, %bx
    1       0   1004 test $0, %bx
    1       0   1005 jgt .top
    1       0   1006 halt
    1       1   ----- Halt;Switch -----  ----- Halt;Switch -----
    1       1                            1000 mov 2000, %ax
    1       1                            1001 add $1, %ax
    2       1                            1002 mov %ax, 2000
    2       0                            1003 sub  $1, %bx
    2       0                            1004 test $0, %bx
    2       0                            1005 jgt .top
    2       0                            1006 halt

Here you can see each thread ran once, and each updated the shared
variable at address 2000 once, thus resulting in a count of two there.

The "Halt;Switch" line is inserted whenever a thread halts and another
thread must be run.

One last example: run the same thing above, but with a smaller interrupt
frequency. Here is what that will look like:

[mac Race-Analyze] ./x86.py -p looping-race-nolock.s -t 2 -a bx=1 -M 2000 -i 2

 2000          Thread 0                Thread 1
    ?
    ?   1000 mov 2000, %ax
    ?   1001 add $1, %ax
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?                            1000 mov 2000, %ax
    ?                            1001 add $1, %ax
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?   1002 mov %ax, 2000
    ?   1003 sub  $1, %bx
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?                            1002 mov %ax, 2000
    ?                            1003 sub  $1, %bx
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?   1004 test $0, %bx
    ?   1005 jgt .top
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?                            1004 test $0, %bx
    ?                            1005 jgt .top
    ?   ------ Interrupt ------  ------ Interrupt ------
    ?   1006 halt
    ?   ----- Halt;Switch -----  ----- Halt;Switch -----
    ?                            1006 halt

As you can see, each thread is interrupt every 2 instructions, as we specify
via the "-i 2" flag. What is the value of memory[2000] throughout this run?
What should it have been?

Now let's give a little more information on what can be simulated
with this program. The full set of registers: %ax, %bx, %cx, %dx, and the PC. 
In this version, there is no support for a "stack", nor are there call
and return instructions.

The full set of instructions simulated are:

mov immediate, register     # moves immediate value to register
mov memory, register        # loads from memory into register
mov register, register      # moves value from one register to other
mov register, memory        # stores register contents in memory
mov immediate, memory       # stores immediate value in memory

add immediate, register     # register  = register  + immediate
add register1, register2    # register2 = register2 + register1
sub immediate, register     # register  = register  - immediate
sub register1, register2    # register2 = register2 - register1

neg register                # negates contents of register

test immediate, register    # compare immediate and register (set condition codes)
test register, immediate    # same but register and immediate
test register, register     # same but register and register

jne                         # jump if test'd values are not equal
je                          #                       ... equal
jlt                         #     ... second is less than first
jlte                        #               ... less than or equal
jgt                         #            ... is greater than
jgte                        #               ... greater than or equal

push memory or register     # push value in memory or from reg onto stack
                            # stack is defined by sp register
pop [register]              # pop value off stack (into optional register)
call label                  # call function at label

xchg register, memory       # atomic exchange: 
                            #   put value of register into memory
                            #   return old contents of memory into reg
                            # do both things atomically

yield                       # switch to the next thread in the runqueue

nop                         # no op


Notes: 
- 'immediate' is something of the form $number
- 'memory' is of the form 'number' or '(reg)' or 'number(reg)' or 
   'number(reg,reg)' or 'number(reg,reg,scale)' (as described above)
- 'register' is one of %ax, %bx, %cx, %dx

Finally, here are the full set of options to the simulator are available with
the -h flag: 

Usage: x86.py [options]

Options:
  -s SEED, --seed=SEED  the random seed
  -t NUMTHREADS, --threads=NUMTHREADS
                        number of threads
  -p PROGFILE, --program=PROGFILE
                        source program (in .s)
  -i INTFREQ, --interrupt=INTFREQ
                        interrupt frequency
  -P PROCSCHED, --procsched=PROCSCHED
                        control exactly which thread runs when
  -r, --randints        if interrupts are random
  -a ARGV, --argv=ARGV  comma-separated per-thread args (e.g., ax=1,ax=2 sets
                        thread 0 ax reg to 1 and thread 1 ax reg to 2);
                        specify multiple regs per thread via colon-separated
                        list (e.g., ax=1:bx=2,cx=3 sets thread 0 ax and bx and
                        just cx for thread 1)
  -L LOADADDR, --loadaddr=LOADADDR
                        address where to load code
  -m MEMSIZE, --memsize=MEMSIZE
                        size of address space (KB)
  -M MEMTRACE, --memtrace=MEMTRACE
                        comma-separated list of addrs to trace (e.g.,
                        20000,20001)
  -R REGTRACE, --regtrace=REGTRACE
                        comma-separated list of regs to trace (e.g.,
                        ax,bx,cx,dx)
  -C, --cctrace         should we trace condition codes
  -S, --printstats      print some extra stats
  -v, --verbose         print some extra info
  -H HEADERCOUNT, --headercount=HEADERCOUNT
                        how often to print a row header
  -c, --compute         compute answers for me


Most are obvious. Usage of -r turns on a random interrupter (from 1 to intfreq
as specified by -i), which can make for more fun during homework problems.

-P lets you specify exactly which threads run when; 
   e.g., 11000 would run thread 1 for 2 instructions, then thread 0 for 3,
   then repeat

-L specifies where in the address space to load the code.

-m specified the size of the address space (in KB).

-S prints some extra stats

-c lets you see the values of the traced registers or memory values
   (otherwise they show up as question marks)

-H lets you specify how often to print a row header (useful for long traces)

Now you have the basics in place; read the questions at the end of the chapter
to study this race condition and related issues in more depth.

Questions

  1. Examine flag.s. This code “implements” locking with a single memory flag. Can you understand the assembly?

  2. When you run with the defaults, does flag.s work? Use the -M and -R flags to trace variables and registers (and turn on -c to see their values). Can you predict what value will end up in flag?

  3. Change the value of the register %bx with the -a flag(e.g.,-abx=2,bx=2 if you are running just two threads). What does the code do? How does it change your answer for the question above?

  4. Set bx to a high value for each thread, and then use the -i flag to generate different interrupt frequencies; what values lead to bad outcomes? Which lead to good outcomes?

  5. Now let’s look at the program test-and-set.s. First, try to understand the code, which uses the xchg instruction to build a simple locking primitive. How is the lock acquire written? How about lock release?

  6. Now run the code, changing the value of the interrupt interval (-i) again, and making sure to loop for a number of times. Does the code always work as expected? Does it sometimes lead to an inefficient use of the CPU? How could you quantify that?

  7. Use the -P flag to generate specific tests of the locking code. For example, run a schedule that grabs the lock in the first thread, but then tries to acquire it in the second. Does the right thing happen? What else should you test?

  8. Now let’s look at the code in peterson.s, which implements Peterson’s algorithm (mentioned in a sidebar in the text). Study the code and see if you can make sense of it.

  9. Now run the code with different values of -i. What kinds of different behavior do you see? Make sure to set the thread IDs appropriately (using -a bx=0,bx=1 for example) as the code assumes it.

  10. Can you control the scheduling (with the -P flag) to “prove” that the code works? What are the different cases you should show hold? Think about mutual exclusion and deadlock avoidance.

  11. Now study the code for the ticket lock in ticket.s. Does it match the code in the chapter? Then run with the following flags: -a bx=1000,bx=1000 (causing each thread to loop through the critical section 1000 times). Watch what happens; do the threads spend much time spin-waiting for the lock?

  12. How does the code behave as you add more threads?

  13. Now examine yield.s, in which a yield instruction enables one thread to yield control of the CPU (realistically, this would be an OS primitive, but for the simplicity, we assume an instruction does the task). Find a scenario where test-and-set.s wastes cycles spinning, but yield.s does not. How many instructions are saved? In what scenarios do these savings arise?

  14. Finally, examine test-and-test-and-set.s. What does this lock do? What kind of savings does it introduce as compared to test-and-set.s?

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