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    <p>Under the covers, <code>i += 1</code> does something like this:</p> <pre><code>try: i = i.__iadd__(1) except AttributeError: i = i.__add__(1) </code></pre> <p>While <code>i = i + 1</code> does something like this:</p> <pre><code>i = i.__add__(1) </code></pre> <p>This is a slight oversimplification, but you get the idea: Python gives types a way to handle <code>+=</code> specially, by creating an <code>__iadd__</code> method as well as an <code>__add__</code>.</p> <p>The intention is that mutable types, like <code>list</code>, will mutate themselves in <code>__iadd__</code> (and then return <code>self</code>, unless you're doing something very tricky), while immutable types, like <code>int</code>, will just not implement it.</p> <p>For example:</p> <pre><code>&gt;&gt;&gt; l1 = [] &gt;&gt;&gt; l2 = l1 &gt;&gt;&gt; l1 += [3] &gt;&gt;&gt; l2 [3] </code></pre> <p>Because <code>l2</code> is the same object as <code>l1</code>, and you mutated <code>l1</code>, you also mutated <code>l2</code>.</p> <p>But:</p> <pre><code>&gt;&gt;&gt; l1 = [] &gt;&gt;&gt; l2 = l1 &gt;&gt;&gt; l1 = l1 + [3] &gt;&gt;&gt; l2 [] </code></pre> <p>Here, you didn't mutate <code>l1</code>; instead, you created a new list, <code>l1 + [3]</code>, and rebound the name <code>l1</code> to point at it, leaving <code>l2</code> pointing at the original list.</p> <p>(In the <code>+=</code> version, you were also rebinding <code>l1</code>, it's just that in that case you were rebinding it to the same <code>list</code> it was already bound to, so you can usually ignore that part.)</p>
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    1. POWhen is "i += x" different from "i = i + x" in Python?
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    1. COdoes `__iadd__` actually call `__add__` in the event of an `AttributeError`?
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    2. COWell, `i.__iadd__` doesn't call `__add__`; it's `i += 1` that calls `__add__`.
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    3. COerrr... Yeah, that's what I meant. Interesting. I didn't realize that was done automatically.
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