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    <p>You need to make the <code>enable_if</code> dependent on a template parameter. As it is now, it is only dependent on the outer template parameter within the template definition. But if you instantiate a class from the outer template, then your assignment operator template that is instantiated into that class is not dependent on a template parameter anymore because <code>T</code> will have been substituted already. </p> <p>Just introduce a dummy parameter equal to <code>T</code></p> <pre><code>template&lt;typename T1 = T, typename = typename std::enable_if&lt; ( !std::is_same&lt; X&lt;T1&gt;, typename StrangerTypeRules&lt;T1&gt;::type &gt;::value ), X&lt;T1&gt; &gt;::type &gt; X&lt;T1&gt; &amp; operator=( const typename StrangerTypeRules &lt;T1&gt;::type&amp; rhs ) { return *this; } </code></pre> <p>Using <code>T1</code> in just one of theplaces within the <code>enable_if&lt;...&gt;</code> template arguments would suffice already, because that already makes the <code>enable_if</code> dependent. </p> <p>However it is <em>not</em> the call to your <code>operator=</code> that was previously illformed but the <em>declaration</em> of it, which conflicts with the copy-assignment operator, so the <code>enable_if</code> has little utility here. Just replace your code by</p> <pre><code>template&lt;typename T1 = T&gt; X&lt;T1&gt; &amp; operator=( const typename StrangerTypeRules &lt;T1&gt;::type&amp; rhs ) { return *this; } </code></pre> <p>Since this <code>operator=</code> is a template, it will not conflict with the non-template overload. And also because it is a template, when you call it the compiler will prefer the non-template if <code>T</code> is <code>X&lt;T&gt;</code>.</p>
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    1. USJohannes Schaub - litb
    1. USJohannes Schaub - litb
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    1. COIs `typename StrangerTypeRules <T1>::type` deducible here? I thought this would be a non-deducible context. Or does defaulting `T1` to `T` change that?
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      1. USildjarn
    2. CO@ildjarn it isn't. but it doesn't need to be deduced either, because `T1` has a default argument. After deduction, the default argument is used if `T1` hasn't otherwise been deduced (which will be the case here).
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    3. CO@ildjarn in this case we could have used `T` in the parameter and return type and stick to `= void` for the default argument. But I like to make the parameter and return types as dependent as possible, because this will make the template as unique as possible and allows more overloads. For example if I would have overloaded it and used `const typename MoreStrangeTypeRules<T1>::type&` and it would yield `X<T1>` as parameter type, it would still work. Had I used `T`, it would have failed because both overloads would have `const X<T>&` as parameter type at declaration time.
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