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copied!<p>From the little research I did (and ignoring the c++0x comment from Johannes), my answer is that it depends what you want the <code>enable_if</code> for. If you want the conversion operation to <code>T</code> to exist or not from the type <code>T</code> then it seems that the answer is no, there is no way in C++03 (as Ugo said). But if you need the <code>enable_if</code> to change the <em>behavior</em> of the operator depending on the type of <code>T</code> then yes, there is a workaround which is to call an enabled helper function (called <code>to<T></code> as Matthieu suggested).</p> <pre><code>#include<iostream> #include<boost/utility/enable_if.hpp> #include<boost/type_traits/is_class.hpp> struct B{ B(const B& other){} B(){} }; struct A{ template<class T> T to(typename boost::enable_if_c<not boost::is_class<T>::value, void*>::type = 0){ std::clog << "converted to non class" << std::endl; return T(0); } template<class T> T to(typename boost::enable_if_c<boost::is_class<T>::value, void*>::type = 0){ std::clog << "conveted to class" << std::endl; return T(); } template<class T> operator T(){ return to<T>(); } }; int main(){ A a; double d = (double)a; // output: "converted to non class" B b = (B)(a); // output: "converted to class" return 0; } </code></pre> <p>For the record, I was frustrated with this for several days, until I realized that I wanted <code>enable_if</code> not for SFINAE but for compile-time behavior change. You may also find that this is the real reason for your need for <code>enable_if</code> also. Just a suggestion.</p> <p>(Please note that this is an answer for the C++98 era)</p>

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