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copied!<p>It's a bug, at least in C++03 (not sure about C++11).</p> <p>According to the overload resolution rules in C++03 §13.3.1.2, the built-in <code>||</code> operator is selected, since no user-defined <code>||</code> operator is defined for <code>S</code>.</p> <p>§5.15/1 says:</p> <blockquote> <p>The <code>||</code> operator groups left-to-right. The operands are both implicitly converted to <code>bool</code> (clause 4). [...]</p> </blockquote> <p>§12.3/2 says:</p> <blockquote> <p>User-defined conversions are applied only where they are unambiguous (10.2, 12.3.2). [...]</p> </blockquote> <p>§12.3/5:</p> <blockquote> <p>User-defined conversions are used implicitly only if they are unambiguous. [...] Function overload resolution (13.3.3) selects the best conversion function to perform the conversion.</p> </blockquote> <p>§13.3.2/3:</p> <blockquote> <p>Second, for <code>F</code> to be a viable function, there shall exist for each argument an implicit conversion sequence (13.3.3.1) that converts that argument to the corresponding parameter of <code>F</code>.</p> </blockquote> <p>Clearly <code>S</code> defines a user-defined conversion to <code>bool</code>. The built-in <code>||</code> operator is a viable function for the overload resolution, and since it's the only one, it's the best. So the expression is well-formed.</p> <p>Also of note is §4/3, which says:</p> <blockquote> <p>An expression <code>e</code> can be implicitly converted to a type <code>T</code> if and only if the declaration “<code>T t=e;</code>” is well- formed, for some invented temporary variable <code>t</code> (8.5). [...]</p> </blockquote> <p>So I'm curious whether Visual Studio also produces an error for the statement <code>bool b = S();</code>.</p>

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