I having using python now for about a year and am fairly familiar with it. Though I am quite new to threads and am a little confused by what data threads share.
I have been reading through stuff online which all seem to agree that threads share the same memory space. Though in trying to demonstrate this to myself, it seems I have an incorrect understanding of how this sharing works.
I wrote a short script to just add one to a local variable three times. I create two threads at a time using the same function. I would have thought that due to a shared memory the X variable in one thread would also be increased while it sleeps due to another thread increasing its own X, and vice versa. So after the second loop of thread one where x=2 while thread two sleeps, I would have thought thread two would come out of its sleep with x = 2 and not x = 1. Though as the order of the printed statements suggests, the variables are not shared between the threads.
My question is if you have multiple threads running at once using the same function, will the variables in each thread be kept separate every time throughout the program running (provided no globals are defined)? And then what exactly does this shared memory mean?
Any guidance on the issue (or general threading advice) would be greatly appreciated.
import threading
from time import sleep
def increase(x):
for i in range(3):
print(f"[{threading.currentThread().getName()}] X is {x}")
x = 1
print(f"[{threading.currentThread().getName()}] X is now {x} after increase")
sleep(0.5)
print(f"[{threading.currentThread().getName()}] X is now {x} after sleep")
return x
def main():
x = 0
first = threading.Thread(name="Thread One", target=increase,args=([x]))
second = threading.Thread(name="Thread Two", target=increase,args=([x]))
first.start()
second.start()
first.join()
second.join()
if __name__ == "__main__":
main()
And the result is:
[Thread One] X is 0
[Thread One] X is now 1 after increase
[Thread Two] X is 0
[Thread Two] X is now 1 after increase
[Thread Two] X is now 1 after sleep[Thread One] X is now 1 after sleep
[Thread One] X is 1
[Thread One] X is now 2 after increase
[Thread Two] X is 1
[Thread Two] X is now 2 after increase
[Thread One] X is now 2 after sleep[Thread Two] X is now 2 after sleep
[Thread Two] X is 2
[Thread Two] X is now 3 after increase
[Thread One] X is 2
[Thread One] X is now 3 after increase
[Thread One] X is now 3 after sleep[Thread Two] X is now 3 after sleep
CodePudding user response：
You're correct in saying that memory is shared but the intricacies of it is deeper. What you're getting confused with is immutable vs mutable types. You can find out more here. Ive removed the for loop since it gets confusing:
import threading
from time import sleep
def increase(x):
print(f"[{threading.currentThread().getName()}] address of x: {hex(id(x))} ")
print(f"[{threading.currentThread().getName()}] X is {x}")
x = 1
print(f"[{threading.currentThread().getName()}] address of x after increment: {hex(id(x))} ")
print(f"[{threading.currentThread().getName()}] X is now {x} after increase")
sleep(0.5)
print(f"[{threading.currentThread().getName()}] X is now {x} after sleep")
print(f"[{threading.currentThread().getName()}] address of x after sleep: {hex(id(x))} ")
return x
def main():
x = 0
first = threading.Thread(name="Thread One", target=increase, args=([x]))
second = threading.Thread(name="Thread Two", target=increase, args=([x]))
first.start()
second.start()
first.join()
second.join()
if __name__ == "__main__":
main()
What ive done here is printed the address of x in the threads. The output:
[Thread One] address of x: 0x7ffbbebb7c20
[Thread One] X is 0
[Thread One] address of x after increment: 0x7ffbbebb7c40
[Thread One] X is now 1 after increase
[Thread Two] address of x: 0x7ffbbebb7c20
[Thread Two] X is 0
[Thread Two] address of x after increment: 0x7ffbbebb7c40
[Thread Two] X is now 1 after increase
[Thread Two] X is now 1 after sleep[Thread One] X is now 1 after sleep
[Thread One] address of x after sleep: 0x7ffbbebb7c40
[Thread Two] address of x after sleep: 0x7ffbbebb7c40
You will notice that the first print line when im just reading x the address is 0x7ffbbebb7c20 after updating it thread 1 and 2 get a different addresses: 0x7ffbbebb7c40. Now they both get the same address cause python tries to keeg the memory footprint lower. You can find more about that here But for our purposes the function gets the same variable to read and once you try write or update that variable a copy of it is made for that thread. This only happens if you're working with a immutable type (int, string, instances etc.) if you pass a mutable type like dict:
import threading
from time import sleep
def increase(test_var):
print(f"[{threading.currentThread().getName()}] Address of test_var: {hex(id(test_var))}")
print(f"[{threading.currentThread().getName()}] Address of test_var['key']: {hex(id(test_var['key']))}")
print(f"[{threading.currentThread().getName()}] test_var['key'] is {test_var['key']}")
test_var['key'] = 1
print(f"[{threading.currentThread().getName()}] test_var['key'] is now {test_var['key']} after increase")
print(f"[{threading.currentThread().getName()}] Address of test_var after increment: {hex(id(test_var))}")
print(f"[{threading.currentThread().getName()}] Address of test_var['key'] after increment: {hex(id(test_var['key']))}")
sleep(0.5)
print(f"[{threading.currentThread().getName()}] test_var['key'] is now {test_var['key']} after sleep")
print(f"[{threading.currentThread().getName()}] Address of test_var after sleep: {hex(id(test_var))}")
print(f"[{threading.currentThread().getName()}] Address of test_var['key'] after sleep: {hex(id(test_var['key']))}")
return test_var
def main():
test_var = {'key': 0}
first = threading.Thread(name="Thread One", target=increase, args=([test_var]))
second = threading.Thread(name="Thread Two", target=increase, args=([test_var]))
first.start()
second.start()
first.join()
second.join()
if __name__ == "__main__":
main()
The output is what you expected:
[Thread One] Address of test_var: 0x22216509a98
[Thread One] Address of test_var['key']: 0x7ffbaf7d7c20
[Thread One] test_var['key'] is 0
[Thread One] test_var['key'] is now 1 after increase
[Thread One] Address of test_var after increment: 0x22216509a98
[Thread One] Address of test_var['key'] after increment: 0x7ffbaf7d7c40
[Thread Two] Address of test_var: 0x22216509a98
[Thread Two] Address of test_var['key']: 0x7ffbaf7d7c40
[Thread Two] test_var['key'] is 1
[Thread Two] test_var['key'] is now 2 after increase
[Thread Two] Address of test_var after increment: 0x22216509a98
[Thread Two] Address of test_var['key'] after increment: 0x7ffbaf7d7c60
[Thread Two] test_var['key'] is now 2 after sleep
[Thread Two] Address of test_var after sleep: 0x22216509a98
[Thread Two] Address of test_var['key'] after sleep: 0x7ffbaf7d7c60
[Thread One] test_var['key'] is now 2 after sleep
[Thread One] Address of test_var after sleep: 0x22216509a98
[Thread One] Address of test_var['key'] after sleep: 0x7ffbaf7d7c60
Notice how the address of test_var (0x22216509a98) doesn't change between the threads Because its mutable and can be shared across threads.
CodePudding user response：
In your case, x is shared by a function parameter as a copy and not a reference. If you want to increase your counter you have to encapsulate it in a class.
For ex:
import threading
from time import sleep
class foo:
x = 0
def increase(foo):
for i in range(3):
print(f"[{threading.currentThread().getName()}] X is {foo.x}")
foo.x = 1
print(f"[{threading.currentThread().getName()}] X is now {foo.x} after increase")
sleep(0.5)
print(f"[{threading.currentThread().getName()}] X is now {foo.x} after sleep")
return foo.x
def main():
x = foo()
first = threading.Thread(name="Thread One", target=increase,args=([x]))
second = threading.Thread(name="Thread Two", target=increase,args=([x]))
first.start()
second.start()
first.join()
second.join()
if __name__ == "__main__":
main()
Note: Python threads are kind of specific. You can have a look at this video https://www.youtube.com/watch?v=Obt-vMVdM8s
------------ Edit -------------
To be more precise. In your case, x is an int and so it is copied on each function call. Same behaviour whether it's a string or float.
You will have the same behaviour without thread:
def increase(x):
for i in range(3):
print(x)
x = 1
return x
x = 0
increase(x)
assert x == 0
x = 1
increase(x)
assert x == 1
CodePudding user response：
The answer that you accepted does not directly answer this question:
if you have multiple threads running at once using the same function, will the variables in each thread be kept separate?
Every time a function is called, a new activation record is created, and the argument values and all of the local variable values are stored in that record. When the function returns, the activation record is destroyed.
That means, that the x argument in your increase(x) function is a different variable in each call to the function. If a function calls itself recursively, then the args and locals are different variables in each recursive call, and if the function is called in multiple threads, then the args and locals are different variables in each of the threads.
I have been reading through stuff online which all seem to agree that threads share the same memory space.
Absolutely true, but an argument or a local is not a definite location in memory. A global is a definite location in memory. So, if you have some global g, every thread will agree that g has the same value. And, a Python object, so long as it exists, occupies a definite location, so every thread that has a reference to the same object will see it in the same state. But, a local variable, as described above, occupies a different memory location in each activation of the function that declares it.
In a nutshell, "local" doesn't just mean local to this function, it means local to this function call.