问题描述:

Is sequence unpacking atomic? e.g.:

`(a, b) = (c, d)`

I'm under the impression it is not.

Edit: I meant atomicity in the context of multi-threading, i.e. whether the entire statement is indivisible, as atoms used to be.

It is one operation; the right-hand expression is evaluated before the left-hand assignment is applied:

```
>>> a, b = 10, 20
>>> a, b
(10, 20)
>>> b, a = a, b
>>> a, b
(20, 10)
>>> a, b = a*b, a/b
>>> a, b
(200, 2)
```

Or, if you are talking about multi-threaded environments, then the assignment is *not* atomic; the interpreter evaluates a tuple assignment with a single opcode, but uses separate opcodes to then store the results into each affected variable:

```
>>> def t(self): a,b=20,20
...
>>> dis.dis(t)
1 0 LOAD_CONST 2 ((20, 20))
3 UNPACK_SEQUENCE 2
6 STORE_FAST 1 (a)
9 STORE_FAST 2 (b)
12 LOAD_CONST 0 (None)
15 RETURN_VALUE
```

However, *normal* assigment is always going to be at least two opcodes (one for the right-hand expression, one for storing the result), so in python *in general* assigment is not atomic. Sequence unpacking is no different.

Definitely not atomic in a multi-threaded environment, tested using the following script:

```
import threading
a, b = 10, 10
finished = False
def thr():
global finished
while True:
# if sequence unpacking and assignment is atomic then (a, b) will always
# be either (10, 10) or (20, 20). Could also just check for a != b
if (a, b) in [(10, 20), (20, 10)]:
print('Not atomic')
finished = True
break
t = threading.Thread(target=thr)
t.start()
while True:
for i in range(1000000):
a, b = 20, 20
a, b = 10, 10
if finished:
t.join()
break
```

Tested using CPython 2.6, 2.7, and 3.2. On each version this script printed 'Not atomic' and exited in well under a second.