[Note: at the time this post was written (September 2016), the version of std::variant proposed for C++17 permitted reference types at alternatives, although the semantics of assignment were not clear. At the November 2016 meeting the C++ commitee chose to resolve this by forbidding variant<T&> . This removes the inconsistency with std::optional<T&> , and thus the main point I was trying to make in this post. Nonetheless, I’d still like to see optional<T&> and I hope we can come to some consensus about the expected assignment semantics at some point in the future.]

I have a confession to make. Whenever I’ve come across code that looks like this:

struct example { example () = default ; example ( std :: string & s ) : str_ { s } {} private: boost :: optional < std :: string &> str_ {}; };

there is a little voice inside my head that whispers “why didn’t you just use a pointer?”. Like so, for instance:

struct example { example () = default ; example ( std :: string & s ) : str_ { & s } {} private: std :: string * str_ = nullptr ; };

This is equivalent to the first example, except that it’s slightly less typing, it doesn’t have any dependencies, and feels in some sense “cleaner”. I personally have always preferred it.

Except, I was wrong. After attending Bjarne Stroustrup’s keynote and this excellent talk at Cppcon this morning, I’m persuaded that optional references are a good thing. In this post I hope to be able to convince you of the same.

std::optional<T&>?

It seems I’m not the only one who preferred the pointer form. Whereas boost::optional allows reference types, the standardised version std::optional (forthcoming in C++17, and available now in libstdc++ and libc++ as std::experimental::optional ) does not. N3527, revision 3 of the paper proposing std::optional , says:

The intention is that the Committee should have an option to accept optional values without optional references, if it finds the latter concept inacceptable [sic]

The paper also goes on to talk about one potential pitfall of optional<T&> , namely assignment, that we’ll talk about later. But first, the pros.

Expressing intent

Pointers represent memory addresses. When we say

int a = 0 ; int * b = & a ;

we are assigning the address of a to the variable b . But if we’re using a pointer as an optional reference then this is misleading: we don’t care about memory addresses at all. We only care if it’s nullptr or not. Compare the above with

int a = 0 ; optional < int &> b { a };

Here, we are explicitly saying that b is either empty, or a reference to an int . The difference becomes more clear when we look at a function signature:

void f ( int * pi );

Is pi allowed to be nullptr here? Without looking at the function documentation, we have no idea. Contrast this with

void g ( optional < int &> oi );

Now it is immediately clear that g() takes a reference to an int , or nothing. We are expressing our intent much more clearly.

Bjarne made the point in his keynote (as he has done several times before) that he would like C++ to be more friendly to novice programmers. By introducing std::array<T> , the committee removed one of the major uses of raw pointers inherited from C; permitting optional<T&> would remove another. It would allow us to say “only use raw pointers in low-level code”, and would make the learning and teaching of C++ that little bit easier.

Better syntax

I don’t think I’m alone in finding the syntax for declaring const pointers and pointers-to-const to be anything but intuitive. Any introduction to pointers contains an example such as the following:

int i = 0 ; int * a = & i ; // mutable pointer to mutable int const int * a = & i ; // mutable pointer to const int int const * b = & i ; // as above, pointer to const int int * const c = & i ; // const pointer to mutable int int const * const d = & i ; // const pointer to const int;

(It’s telling that even after many years of writing C and C++, I still had to double-check that to make sure I got it right.)

Compare this to the equivalent with optional references:

int i = 0 ; auto a = optional < int &> { i }; auto b = optional < const int &> { i }; const auto c = optional < int &> { i }; const auto d = optional < const int &> { i };

The use of the template syntax (and AAA style) immediately makes it clear whether it is the wrapper or the contained referee that is const . The same cannot be said for int const * versus int * const , especially for novices.

Again, there is an analogy with std::array vs C arrays: have you ever tried to pass a C array by reference to a function? The syntax is odd to say the least:

void f ( const std :: string ( & arg )[ 4 ]); // huh? are we taking the address of arg?

whereas with std::array , it’s obvious what’s going on:

void f ( const std :: array < std :: string , 4 >& arg ); // oh, I see

As above, clear, consistent, familiar syntax allows us to express our intent more clearly.

No pointer arithmetic

Optional references do not allow pointer arithmetic. This removes a potential source of errors:

int * maybe_increment ( int * pi ) { if ( pi ) { pi += 1 ; } return pi ; }

Did you spot the bug? The above is legal C++ and compiles without warning, but it’s very unlikely that it does what the author intended. Compare this to the optional reference case:

optional < int &> maybe_increment ( optional < int &> oi ) { if ( oi ) { // oi += 1; // Error, no match for operator+= * oi += 1 ; // That's better } return oi ; }

Consistency with std::variant

Another new addition to C++17 is std::variant . One of the things that variant allows is to specify that an empty state is permitted by using std::monostate . For example:

auto v = variant < monostate , int > {}; // v is either empty, or contains an int

Such a variant is conceptually the same as an optional<int> , except that the latter has a more specialised, friendlier API. This is in much the same way that pair<int, float> is conceptually the same as tuple<int, float> , except that it provides a more specialised API to directly access the two members.

But here’s the thing: variant<monostate, int&> is permitted, behaving as if it contained a std::reference_wrapper<int> . So why not optional<int&> ? Such an inconsistency is needless and confusing.

EDIT: /u/tvaneerd pointed out on Reddit that I was mistaken – the standard says it’s permissible to use a reference wrapper, not necessarily std::reference_wrapper . I believe my point stands though: if variant<monostate, T&> is permitted (and it seems that this is explicitly so), then it is inconsistent to forbid optional<T&> .

Generic programming

The lack of optional references means that generic code which deals with optionals is unnecessarily complicated.

N3527, referenced above, acknowledges this, saying:

Users that in generic contexts require to also store optional lvalue references can achieve this effect, even without direct support for optional references, with a bit of meta-programming.

and providing the following workaround:

template < class T > struct generic { typedef T type ; }; template < class U > struct generic < U &> { typedef std :: reference_wrapper < U > type ; }; template < class T > using Generic = typename generic < T >:: type ; template < class X > void generic_fun () { std :: optional < Generic < X >> op ; // ... }

Urgh. This should not be necessary.

Zero cost

Like std::unique_ptr , optional references can be implemented as a zero-overhead abstraction around a raw pointer. In fact, boost::optional does just this.

There is no reason std::optional<T&> could not do the same.

The assignment problem (and what to do about it)

Having given the plus points of allowing optional references, it’s only fair to give some time to the potential downsides as well. In particular, deciding what to do about assignment appears to be a bit tricky.

Version 2 of the paper proposing std::optional (N3406) goes into a bit more detail about the assignment problem. In short, when assigning one optional reference to another, should it behave like a pointer, or like a built-in language reference?

int i = 0 ; int j = 1 ; int * p1 = & i ; int * p2 = & j ; p2 = p1 ; // Pointer is rebound, p2 now points to i int & r1 = i ; int & r2 = j ; r2 = r1 ; // No reference rebinding, j now equals 0 auto o1 = optional < int &> { i }; auto o2 = optional < int &> { j }; o2 = o1 ; // Should we rebind o2, or copy the value?

From what I can gather, this was one of the major points of contention regarding optional references. As the paper makes clear, there are arguments to be made both ways about how it should behave, and it was not obvious as the time the paper was proposed which way optional should go.

However, I believe that we should now firmly come down on the side of rebinding. Why? Recall from above that optional<T> can be regarded as a special case of variant<monostate, T> . So what does variant<monostate, T&> do when we assign to it? I don’t have an implementation of std::variant to hand, but since it permits implementations to store references in a reference_wrapper , ~ I believe it must rebind. ~

EDIT: As above, it’s been pointed out that I jumped to conclusions. The standard merely says that a reference wrapper is permitted, not that this should be std::reference_wrapper . It seems the rebind vs copy-through debate may still be up in the air for variant .

So, whether the committee intended it or not, it has made a decision; variant permits references, and so whatever it does on assignment, optional should do too. (For what it’s worth, boost::optional rebinds on assignment.)

In summary