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    <p>When you do <code>a + b</code>, it calls call <code>self.value.__add__(other)</code>. Since <code>a.value</code> is 10, this calls <code>(10).__add__(b)</code>. Since <code>b</code> is an <code>adder</code> instance, the integer 10 does not know how to add itself to <code>b</code>, so it returns <code>NotImplemented</code>. The <code>__radd__</code> method of <code>b</code> is not called at this stage! Since you return the result of <code>self.value.__add__(other)</code>, your <code>__add__</code> method also returns NotImplemented. Once this happens, <code>b.__radd__(a)</code> is called, and also returns <code>NotImplemented</code> for the same reason --- it delegates to <code>(12).__radd__(a)</code>, which fails. Since both return <code>NotImplemented</code>, Python thinks the objects don't know how to add to each other, so it raises the error.</p> <p>The solution is, as you found, to use <code>+</code> instead of directly calling the magic methods. (You don't need to make a separate class for that; just change your <code>adder</code> class to call <code>self.value + other</code> instead of <code>self.value__add__(other)</code>, etc.) When you use <code>+</code>, you activate Python's internal machinery, whereby it calls <code>__add__</code> on the left operand and then, if that doesn't work, calls <code>__radd__</code> on the right operand. When you directly call <code>__add__</code> and <code>__radd__</code>, you are bypassing this internal machinery and only calling one of the two. This means that you never get to a state where both objects "reduce" themselves to their underlying values. Instead, you try the left-hand <code>__add__</code>, and when it fails, you don't "fall back" to <code>__radd__</code> --- you start again from the beginning with <code>__radd__</code>, losing the "value reduction" you did in the first part.</p> <p>The crucial thing is that <code>10 + b</code> knows to call <code>b.__radd__(10)</code> if <code>(10).__add__(b)</code> fails, but if you just directly call <code>(10).__add__(b)</code>, it doesn't know anything about <code>__radd__</code>, so <code>b.__radd__(10)</code> is never called.</p>
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