The semantics of assignments in Python are not the same as assignment in C. When you assign a local like `x = some_expression` in Python, you can read it as, “E

The semantics of assignments in Python are not the same as assignment in C. When you assign a local like `x = some_expression` in Python, you can read it as, “Evaluate `some_expression` now, and call that result `x` in this local namespace.” The behavior that results from your example follows from this rule. First, evaluate `1` and call it `sheep`. Then evaluate whatever `sheep` is, once, to get `1` (the same object in memory as every other literal `1` in Python) and call it `goats`. The last line is where the rule matters: The statement `sheep = sheep + 10` can be read as, “Evaluate `sheep + 10` and call the result `sheep`.” The statement reassigns the name `sheep` in the local namespace to point to a different object, one created by evaluating `sheep + 10`. The actual memory location that `sheep` referred to previously (containing the `int` object `1`) is not changed at all — assignment to a local will never change the value of any other local. This is easy to remember if you recall that a local namespace is effectively just a `dict`. Your example is equivalent to: d = {} d["sheep"] = 1 d["goats"] = d["sheep"] d["sheep"] = d["sheep"] + 10 It should be clear even to beginners that `d["goats"]` has a final value of `1`, not `11`, because the right-hand side of `d["goats"] = d["sheep"]` is only evaluated once, and at that time it evaluates to `1`. Assignment using locals behaves in exactly the same way.