* Conversion explicit conversion to bool should be declared as explicit.
Doesn't an explicit conversion disallows to use it in a conditional expression as in if(opt) {}
I believe that the primary motivation for the explicit conversion operators was to provide a simpler alternative for safe-bool idiom, and address cases like this one. http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2007/n2437.pdf
* Should exception safety guarantees for assignment operations be enforced? In boost::optional they only provide a basic guarantee. One option is to provide strong guarantee for types T which implement operator= with strong guarantee, and give basic guarantee for other cases Other option (not sure if it is possible) is to provide strong guarantee for all cases at the cost of run-time performance (heap allocation).
Provide the same exception guaranties as the T assignment seems the more reasonable.
My apologies, I wrote this in a hurry an didn't make myself clear. There are three cases to consider (leaving move assignment and no-throw copyable types aside): 1. Our stored type T provides a strong assignment - the choice is obvious: optional<T>'s assignment is also strong 2. T has "basic" assignment - the natural choice is to provide also a basic guarantee for optional<T>'s assignment. 3. T is neither copyable nor moveable -- it is like a scope guard, or boost::scoped_ptr. In this case optional<T> still provides assignment by making a "pseudo destructor call" followed by a call to placement new. -- but this gives only basic guarantee. The question is if for this third variant we should be happy with only basic guarantee or to use a trick boost::variant is using with auxiliary heap-allocated object? (I do not even know if this is applicable though).
template< typename ... Args> optional( Args&& ...args ); I like them. The constructor will clash with another optional constructor: optional<T>::optional( bool cond, const T& val );
I don't understand why.
Currently boost::optional provides this constructor: optional<T>::optional( bool cond, const T& val ); If cond is true the optional is initialized with object val, otherwise it is uninitialized. Now suppose I have a following type: struct MyType { bool on; int m1; int m2; MyType( bool on, MyType const& rhs ) :on{on}, m1{rhs.m1}, m2{rhs.m2} {} }; MyType mt { true, 1, 2 }; boost::optional<MyType> o1{ true, mt }; // optional initialized with mt (copy ctor used) boost::optional<MyType> o2{ boost::in_place(true, mt) }; // uses 2-arg ctor But if std::tr2::optional has "variadic" constructor: std::tr2::optional<MyType> mt{ true, mt }; // uses 2-arg ctor Regards, &rzej