Using a global variable with a thread

Question

How do I share a global variable with thread   My Python code example is   from threading import Thread import time a   0   global variable  def thread1 threadname        read variable  a  modify by thread 2  def thread2 threadname       while 1          a    1         time sleep 1   thread1   Thread  target thread1  args   Thread-1       thread2   Thread  target thread2  args   Thread-2        thread1 join   thread2 join     I don t know how to get the two threads to share one variable

User · Accepted Answer

You just need to declare a as a global in thread2  so that you aren t modifying an a that is local to that function   def thread2 threadname       global a     while True          a    1         time sleep 1    In thread1  you don t need to do anything special  as long as you don t try to modify the value of a  which would create a local variable that shadows the global one  use global a if you need to    def thread1 threadname        global a         Optional if you treat a as read-only     while a  lt  10          print a

User · Answer

A lock should be considered to use  such as threading Lock  See lock-objects for more info   The accepted answer CAN print 10 by thread1  which is not what you want  You can run the following code to understand the bug more easily   def thread1 threadname       while True        if a   2 and not a   2            print  unreachable    def thread2 threadname       global a     while True          a    1   Using a lock can forbid changing of a while reading more than one time   def thread1 threadname       while True        lock a acquire         if a   2 and not a   2            print  unreachable         lock a release    def thread2 threadname       global a     while True          lock a acquire           a    1         lock a release     If thread using the variable for long time  coping it to a local variable first is a good choice

User · Answer

Thanks so much Jason Pan for suggesting that method  The thread1 if statement is not atomic  so that while that statement executes  it s possible for thread2 to intrude on thread1  allowing non-reachable code to be reached  I ve organized ideas from the prior posts into a complete demonstration program  below  that I ran with Python 2 7   With some thoughtful analysis I m sure we could gain further insight  but for now I think it s important to demonstrate what happens when non-atomic behavior meets threading     ThreadTest01 py - Demonstrates that if non-atomic actions on   global variables are protected  task can intrude on each other  from threading import Thread import time    global variable a   0  NN   100  def thread1 threadname       while True        if a   2 and not a   2            print  unreachable          end of thread1  def thread2 threadname       global a     for   in range NN           a    1         time sleep 0 1        end of thread2  thread1   Thread target thread1  args   Thread1     thread2   Thread target thread2  args   Thread2      thread1 start   thread2 start    thread2 join     end of ThreadTest01 py   As predicted  in running the example  the  unreachable  code sometimes is actually reached  producing output   Just to add  when I inserted a lock acquire release pair into thread1 I found that the probability of having the  unreachable  message print was greatly reduced  To see the message I reduced the sleep time to 0 01 sec and increased NN to 1000    With a lock acquire release pair in thread1 I didn t expect to see the message at all  but it s there  After I inserted a lock acquire release pair also into thread2  the message no longer appeared  In hind signt  the increment statement in thread2 probably also is non-atomic

User · Answer

Well  running example   WARNING  NEVER DO THIS AT HOME WORK  Only in classroom      Use semaphores  shared variables  etc  to avoid rush conditions   from threading import Thread import time  a   0    global variable   def thread1 threadname       global a     for k in range 100           print         format threadname  a           time sleep 0 1          if k    5              a    100   def thread2 threadname       global a     for k in range 10           a    1         time sleep 0 2    thread1   Thread target thread1  args   Thread-1     thread2   Thread target thread2  args   Thread-2      thread1 start   thread2 start    thread1 join   thread2 join     and the output   Thread-1 0 Thread-1 1 Thread-1 2 Thread-1 2 Thread-1 3 Thread-1 3 Thread-1 104 Thread-1 104 Thread-1 105 Thread-1 105 Thread-1 106 Thread-1 106 Thread-1 107 Thread-1 107 Thread-1 108 Thread-1 108 Thread-1 109 Thread-1 109 Thread-1 110 Thread-1 110 Thread-1 110 Thread-1 110 Thread-1 110 Thread-1 110 Thread-1 110 Thread-1 110   If the timing were right  the a    100 operation would be skipped   Processor executes at T a 100 and gets 104  But it stops  and jumps to next thread Here  At T 1 executes a 1 with old value of a  a    4   So it computes 5  Jump back  at T 2   thread 1  and write a 104 in memory  Now back at thread 2  time is T 3 and write a 5 in memory  Voila  The next print instruction will print 5 instead of 104    VERY nasty bug to be reproduced and caught

User · Answer

In a function   a    1   will be interpreted by the compiler as assign to a   gt  Create local variable a  which is not what you want  It will probably fail with a a not initialized error since the  local  a has indeed not been initialized    gt  gt  gt  a   1  gt  gt  gt  def f            a    1       gt  gt  gt  f   Traceback  most recent call last     File   lt stdin gt    line 1  in  lt module gt    File   lt stdin gt    line 2  in f UnboundLocalError  local variable  a  referenced before assignment   You might get what you want with the  very frowned upon  and for good reasons  global keyword  like so    gt  gt  gt  def f            global a         a    1       gt  gt  gt  a 1  gt  gt  gt  f    gt  gt  gt  a 2   In general however  you should avoid using global variables which become extremely quickly out of hand  And this is especially true for multithreaded programs  where you don t have any synchronization mechanism for your thread1 to know when a has been modified  In short  threads are complicated  and you cannot expect to have an intuitive understanding of the order in which events are happening when two  or more  threads work on the same value  The language  compiler  OS  processor    can ALL play a role  and decide to modify the order of operations for speed  practicality or any other reason   The proper way for this kind of thing is to use Python sharing tools  locks  and friends   or better  communicate data via a Queue instead of sharing it  e g  like this   from threading import Thread from queue import Queue import time  def thread1 threadname  q        read variable  a  modify by thread 2     while True          a   q get           if a is None  return   Poison pill         print a  def thread2 threadname  q       a   0     for   in xrange 10           a    1         q put a          time sleep 1      q put None    Poison pill  queue   Queue   thread1   Thread  target thread1  args   Thread-1   queue    thread2   Thread  target thread2  args   Thread-2   queue     thread1 start   thread2 start   thread1 join   thread2 join

[python] Using a global variable with a thread

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