Is the safe-bool idiom obsolete in C++11?
问题 This answer of @R. Martinho Fernandes shows, that the safe-bool idiom is apperently deprecated in C++11, as it can be replaced by a simple explicit operator bool() const; according to the standard quote in the answer §4 [conv] p3 : An expression e can be implicitly converted to a type T if and only if the declaration T t=e; is well-formed, for some invented temporary variable t (§8.5). Certain language constructs require that an expression be converted to a Boolean value. An expression e