How to prove that parameter evaluation is “left to right” in Python?

Question

For example, in JavaScript we could write a program like this:

var a = 1;
testFunction(++a, ++a, a);
function testFunc         


        

Answer 1

Using Python 3:

>>> a = []
>>> f = print(
    a.append(1), a[:],
    a.append(2), a[:],
    a.append(3), a[:]
)
None [1] None [1, 2] None [1, 2, 3]

---

Archive:

>>> a = []
>>> f = print(a.append(1), a, a.append(2), a, a.append(3), a)

Curiously enough (at first), this code produces:

None [1, 2, 3] None [1, 2, 3] None [1, 2, 3]

However, dis(f) makes this clearer:

>>> dis(f)

  1           0 LOAD_NAME                0 (print) #Loads the value of 'print' into memory. Precisely, the value is pushed to the TOS (Top of Stack)
    -->       3 LOAD_NAME                1 (a) #Loads the value of object 'a' 
              6 LOAD_ATTR                2 (append) #Loads the append attribute (in this case method)
              9 LOAD_CONST               0 (1) #Loads the constant 1
             12 CALL_FUNCTION            1 #a.append(1) is called
             15 LOAD_NAME                1 (a) #for print(...,a,...)
             18 LOAD_NAME                1 (a) #for the next a.append()
             21 LOAD_ATTR                2 (append) 
             24 LOAD_CONST               1 (2) 
             27 CALL_FUNCTION            1 #a.append(2)
             30 LOAD_NAME                1 (a) 
             33 LOAD_NAME                1 (a) 
             36 LOAD_ATTR                2 (append) 
             39 LOAD_CONST               2 (3) 
             42 CALL_FUNCTION            1 #a.append(3)
             45 LOAD_NAME                1 (a) #loads a to be used thrice by print
             48 CALL_FUNCTION            6 #calls print
             51 PRINT_EXPR                 #prints TOS and clears it
             52 LOAD_CONST               3 (None) #Loads None
             55 RETURN_VALUE             #Returns None

The output of dis(f) is what we expected - L-to-R evaluation. Essentially, this "discrepancy" is a consequence of print() being evaluated last. By then, the value of a has changed to [1, 2, 3] and the same final object is printed thrice.

If we replace a with a[:], we get the expected result.