What is the order of evaluation in python when using pop(), list[-1] and +=?

Question

a = [1, 2, 3] a[-1] += a.pop() 

This results in [1, 6].

a = [1, 2, 3] a[0] += a.pop() 

This results in [4, 2]. What order of evaluation gives these two results?

Answer 1

RHS first and then LHS. And at any side, the evaluation order is left to right.

a[-1] += a.pop() is same as, a[-1] = a[-1] + a.pop()

a = [1,2,3] a[-1] = a[-1] + a.pop() # a = [1, 6] 

See how the behavior changes when we change the order of the operations at RHS,

a = [1,2,3] a[-1] = a.pop() + a[-1] # a = [1, 5] 

Answer 2

The key insight is that a[-1] += a.pop() is syntactic sugar for a[-1] = a[-1] + a.pop(). This holds true because += is being applied to an immutable object (an int here) rather than a mutable object (relevant question here).

The right hand side (RHS) is evaluated first. On the RHS: equivalent syntax is a[-1] + a.pop(). First, a[-1] gets the last value 3. Second, a.pop() returns 3. 3 + 3 is 6.

On the Left hand side (LHS), a is now [1,2] due to the in-place mutation already applied by list.pop() and so the value of a[-1] is changed from 2 to 6.