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POOverloading conversion operator template
 

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    <p>Consider the following simple example</p> <pre><code>struct C { template &lt;typename T&gt; operator T () {return 0.5;} operator int () {return 1;} operator bool () {return false;} }; int main () { C c; double x = c; std::cout &lt;&lt; x &lt;&lt; std::endl; } </code></pre> <p>When compiled with Clang, it gives the following error</p> <pre><code>test.cpp:11:12: error: conversion from 'C' to 'double' is ambiguous double x = c; ^ ~ test.cpp:4:5: note: candidate function operator int () {return 1;} ^ test.cpp:5:5: note: candidate function operator bool () {return false;} ^ test.cpp:3:27: note: candidate function [with T = double] template &lt;typename T&gt; operator T () {return 0.5;} ^ 1 error generated. </code></pre> <p>Other compilers generate similar errors, e.g., GCC and Intel iclc</p> <p>If I remove <code>operator int</code> and <code>operator bool</code>. It compiles fine and work as expected. If only remove one of them, that is keep the template operator and say <code>operator int</code>, then the non-template version is always chosen.</p> <p>My understanding is that only when the template and non-template overloaded functions are equal in the sense that they are both perfect match or both require the same conversion sequence, the non-template version will be preferred. However in this case, it appears that the compiler does not see the operator template as a perfect match. And when both the <code>bool</code> and <code>int</code> overloading are present, then naturally it considers they are ambiguous.</p> <p>In summary, my question is that why the operator template is not considered a perfect match in this case?</p>
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