Python print function does not return the updated variable

Question

I am learning Python and something related to print confused me. Not sure if someone asked the same question already:

>>> x = 1
>>> x, y = x + 2, print(x)
1

I understand that output 1 is the side effect of Python's print function. But why doesn't it print 3 ? I was expecting 3 because x is updated in the second line? I was thinking it is equivalent to (apparently, wrong):

>>> x = 1
>>> x = x + 2
>>> x
3
>>> y = print(x)
3

I would like to understand the logic behind this print function. Why doesn't it print the updated x value?

I am new to the programming world, so any insight is highly appreciated!

Answer 1

Everything on the right-hand side is evaluated first. You can use the python byte-code dissasembler to see what is happening:

>>> import dis
>>> dis.dis('x, y = x + 2, print(x)')
  1           0 LOAD_NAME                0 (x)
              2 LOAD_CONST               0 (2)
              4 BINARY_ADD
              6 LOAD_NAME                1 (print)
              8 LOAD_NAME                0 (x)
             10 CALL_FUNCTION            1
             12 ROT_TWO
             14 STORE_NAME               0 (x)
             16 STORE_NAME               2 (y)
             18 LOAD_CONST               1 (None)
             20 RETURN_VALUE
>>>

Note, x + 2 and print(x) are evaluated first . The BINARY_ADD and CALL_FUNCTION occur before the two STORE_NAME s.

Note, you can think of this as the equivalent of building a tuple first,

temp = (x + 2, print(x))

and then simply:

x, y = temp

However, note, according to the dissasembler, no actual intermediate tuple is created. The call stack is used to store the intermediate values. This is a compiler optimization. However, the optimization does not work for tuples greater than length 3, so using 4, you'll see an intermediate tuple is create:

>>> dis.dis('foo, bar, baz, bang  = bang, baz, bar, foo')
  1           0 LOAD_NAME                0 (bang)
              2 LOAD_NAME                1 (baz)
              4 LOAD_NAME                2 (bar)
              6 LOAD_NAME                3 (foo)
              8 BUILD_TUPLE              4
             10 UNPACK_SEQUENCE          4
             12 STORE_NAME               3 (foo)
             14 STORE_NAME               2 (bar)
             16 STORE_NAME               1 (baz)
             18 STORE_NAME               0 (bang)
             20 LOAD_CONST               0 (None)
             22 RETURN_VALUE
>>>

Note the BUILD_TUPLE and UNPACK_SEQUENCE , which is the general way that unpacking works in Python. It's just the compiler optimizes the common cases of two-or-three with the ROT_TWO and ROT_THREE op codes.

Note, since the right-hand side is evaluate first, this allows for the Python swap idiom to work!

x, y = y, x

If this were equivalent to:

x = y
y = x

You would lose the value for x instead of swapping!

Answer 2

>>> x = 1
>>> x, y = x + 2, print(x)
1

print(x) outputs the value stored in x assigned by you

for printing 3 you need to print(y)

as y stores the value of x+2

Answer 3

Because python processes line by line (meaning after the whole line, process it), so x haven't updated yet, so that's why it's not working as expected, but if you do ; (semicolon) it will work, because semicolon ; is basically a separator of lines too, so basically everything after ; is taken as a new line, demo:

>>> x=1
>>> x, y = x + 2,0; print(x)
3
>>> 

Issue is that y is gonna be 0:

>>> y
0
>>> 

Because i do so, you have to have a value,

But if you care, delete it:

>>> x=1
>>> x, y = x + 2,0; print(x)
3
>>> del y
>>> y
Traceback (most recent call last):
  File "<pyshell#12>", line 1, in <module>
    y
NameError: name 'y' is not defined
>>> 

Answer 4

直到左侧的所有内容都完成处理后，赋值左侧的值才会更新。这就是为什么当您尝试在右侧打印时X尚未为3的原因。