In C++, a one-argument constructor can implicitly be used to convert some source value or object into an instance of your class.
\nTo prevent this you can mark your one-argument constructors as explicit. In that case, if you want the conversion you have to explicitly perform the casting.
\nC++11 adds this functionality to conversion operators as well, however, its behavior surprised me.
\nFor example:<\/p>\n
#include <iostream>\r\n\r\nusing namespace std;\r\n\r\nclass Foo\r\n{\r\npublic:\r\n\texplicit operator bool() const { return true; }\r\n};\r\n\r\nint main()\r\n{\r\n\tFoo f;\r\n\tif (f)\r\n\t{\r\n\t\tcout << "f is true" << endl;\r\n\t}\r\n\treturn 0;\r\n}\r\n<\/pre>\nThe output of this piece of code is “f is true”.
\nIn other words, the compiler is implicitly converting f into a Boolean, even though the conversion operator is marked as explicit.
\nThis surprised me. I immediately blamed it on a compiler bug. However, after some more investigation, it turns out the C++ standard has this notion of “contextually converted to bool”<\/strong>.
\nThe C++ standard has a paragraph explaining this behavior, section 4 Standard conversions [conv]<\/p>\n3 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).
\n4 Certain language constructs require that an expression be converted to a Boolean value. An expression e appearing in such a context is said to be contextually converted to bool and is well-formed if and only if the declaration bool t(e); is well-formed, for some invented temporary variable t (8.5).<\/p><\/blockquote>\n