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

Answer 1

Using a thin wrapper around a list with debugging print-statements can be used to show the order of evaluation in your cases:

class Test(object):
    def __init__(self, lst):
        self.lst = lst

    def __getitem__(self, item):
        print('in getitem', self.lst, item)
        return self.lst[item]

    def __setitem__(self, item, value):
        print('in setitem', self.lst, item, value)
        self.lst[item] = value

    def pop(self):
        item = self.lst.pop()
        print('in pop, returning', item)
        return item

When I now run your example:

>>> a = Test([1, 2, 3])
>>> a[-1] += a.pop()
in getitem [1, 2, 3] -1
in pop, returning 3
in setitem [1, 2] -1 6

So it starts by getting the last item, which is 3, then pops the last item which is also 3, adds them and overwrites the last item of your list with 6. So the final list will be [1, 6].

And in your second case:

>>> a = Test([1, 2, 3])
>>> a[0] += a.pop()
in getitem [1, 2, 3] 0
in pop, returning 3
in setitem [1, 2] 0 4

This now takes the first item (1) adds it to the popped value (3) and overwrites the first item with the sum: [4, 2].

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The general order of evaluation is already explained by @Fallen and @tobias_k. This answer just supplements the general principle mentioned there.