This antipattern keeps coming up, so here’s the blog post I can point people to. Today we’re talking about const_iterator . First of all — you know this — const_iterator is different from iterator const , in exactly the same way that const int * is different from int *const .

(I am “west const” for life, but even a west-conster can write the const on the east-hand side when it is pedagogically useful!)

struct MyContainer { using iterator = MyIterator; iterator begin(); // GOOD! const iterator begin() const; // WRONG! };

Remember that const-qualifying the return type of a function that returns by value is never, ever useful; and quite often is a pessimization. See “ const is a contract” (2019-01-03).

Okay, so we all agree: we should be writing our containers like this —

template<bool IsConst> struct MyIterator { // ... }; using Iterator = MyIterator<false>; using ConstIterator = MyIterator<true>; struct MyContainer { using iterator = Iterator; using const_iterator = ConstIterator; iterator begin(); // GOOD! const_iterator begin() const; // GOOD! };

The next pitfall, the one I actually want to talk about, is that now we have two different class types ( MyIterator<false> and MyIterator<true> ), and so we have to think about how they interact with each other. For example, we want this to compile, right?

MyContainer c; MyContainer::const_iterator it = c.begin();

So we need to have an implicit conversion from iterator to const_iterator . And we don’t want to implement it with a conversion operator, because that disables implicit move. (EWG members, take note!) So we break out the converting constructor.

template<bool IsConst> class MyIterator { int *d_; public: // Always permit conversion from this class's own type... MyIterator(const MyIterator<IsConst>& rhs) = default; // ...and always permit conversion FROM the non-const iterator type. MyIterator(const MyIterator<false>& rhs) : d_(rhs.d_) {} };

This doesn’t compile, because in the case that IsConst == false , we’ve just declared the same constructor twice. But if we’re too-clever-for-our-own-good, we simply drop the defaulted copy constructor — it’ll get implicitly generated anyway, right?

template<bool IsConst> class MyIterator { int *d_; public: // Always (implicitly) permit conversion from this class's own type... // ...and always permit conversion FROM the non-const iterator type. MyIterator(const MyIterator<false>& rhs) : d_(rhs.d_) {} }; using Iterator = MyIterator<false>; using ConstIterator = MyIterator<true>;

Now ConstIterator is constructible from either ConstIterator or Iterator , and Iterator is constructible from only Iterator . Ship it, right?

In the case that IsConst == false , the constructor we’ve just provided is the copy constructor. This was intentional; it’s not the pitfall. The pitfall is that, because we provided a curly-braced body for it, it’s no longer trivial. So we’ll find that

static_assert(std::is_copy_constructible_v<ConstIterator>); static_assert(std::is_trivially_copy_constructible_v<ConstIterator>); // GOOD static_assert(std::is_copy_constructible_v<Iterator>); static_assert(not std::is_trivially_copy_constructible_v<Iterator>); // OOPS!

The libc++ implementation of vector<bool>::iterator currently falls into this exact trap. Godbolt:

So if we shouldn’t write the too-clever code above, what should we write? It turns out that all we have to do is be a tiny bit less clever. Thanks to Glen Fernandes for pointing out that this works —

template<bool IsConst> class MyIterator { int *d_; public: template<bool WasConst, class = std::enable_if_t<IsConst || !WasConst>> MyIterator(const MyIterator<WasConst>& rhs) : d_(rhs.d_) {} // OK };

This constructor template is too templatey to be recognized by the compiler as an attempt to define the class’s copy constructor. So the compiler implicitly generates a defaulted (and trivial) copy constructor in both the IsConst and !IsConst cases. Our hand-written constructor gets called only when IsConst && !WasConst . …Which means it might be clearer to replace the || with an && .

We could (and for my money, should) add a declaration for the =default ’ed copy constructor, too.

Personally, I would be inclined to eliminate WasConst and then use my standard metaprogramming trick to “lazify” the evaluation of the enable_if condition — so I would write

template<bool IsConst> class MyIterator { int *d_; public: MyIterator(const MyIterator&) = default; // REDUNDANT BUT GOOD STYLE template<bool IsConst_ = IsConst, class = std::enable_if_t<IsConst_>> MyIterator(const MyIterator<false>& rhs) : d_(rhs.d_) {} // OK };

Read the next section to see one way in which my approach might be considered suboptimal; and the section after that for a way in which it could be considered more optimal.

You might wonder — iterators are cheap to copy (this one is trivial and register-sized), so can we make our converting constructor pass-by-value instead of pass-by-const-reference? It turns out that all non-GCC compilers are unhappy with

template<bool IsConst> class MyIterator { int *d_; public: template<bool IsConst_ = IsConst, class = std::enable_if_t<IsConst_>> MyIterator(MyIterator<false> rhs) : d_(rhs.d_) {} // BOGUS(?) ERROR };

but everyone is happy with

template<bool IsConst> class MyIterator { int *d_; public: template<bool WasConst, class = std::enable_if_t<IsConst || !WasConst>> MyIterator(MyIterator<WasConst> rhs) : d_(rhs.d_) {} // OK(?) };

Finally, if we’re providing a converting constructor from Iterator to ConstIterator , should we also provide a converting assignment operator?

template<bool IsConst> class MyIterator { int *d_; public: MyIterator(const MyIterator&) = default; // REDUNDANT BUT GOOD STYLE MyIterator& operator=(const MyIterator&) = default; // REDUNDANT BUT GOOD STYLE template<bool WasConst, class = std::enable_if_t<IsConst && !WasConst>> MyIterator(const MyIterator<WasConst>& rhs) : d_(rhs.d_) {} template<bool WasConst, class = std::enable_if_t<IsConst && !WasConst>> MyIterator& operator=(const MyIterator<WasConst>& rhs) { d_ = rhs.d_; return *this; } };

Notice that ordinary assignments from Iterator to ConstIterator will compile just fine without such an assignment operator. ConstIterator::operator= is just like any other C++ function; you can call it with any argument that implicitly converts to the type of its formal parameter. And we just got done adding that implicit conversion in the form of our converting constructor.

Iterator it; ConstIterator cit; cit = it; // implicitly convert `it` to ConstIterator, then use the copy assignment operator

However, implicit conversions in C++ are “limit 1 per customer.” The following code using std::reference_wrapper will fail to compile —

Iterator it; ConstIterator cit; cit = std::ref(it);

— unless we add an operator= . And here there is a big difference between

template<bool WasConst, class = std::enable_if_t<IsConst && !WasConst>> MyIterator& operator=(const MyIterator<WasConst>& rhs) { ... } template<bool IsConst_ = IsConst, class = std::enable_if_t<IsConst_>> MyIterator& operator=(const MyIterator<false>& rhs) { ... }

The former requires deducing WasConst , which means it relies on template type deduction, which means that the actual argument must match the formal parameter’s type exactly. Which means that the former operator= will not help us if our goal is to make cit = std::ref(it); compile. We must use the latter version.

So, should we provide this converting operator= ? In my opinion, the right thing to do here is “do as the STL does.” So let’s find out if the STL provides converting assignment operators:

#include <functional> // for reference_wrapper #include <list> std::list<int>::iterator it; std::list<int>::const_iterator cit = std::ref(it); // OK? cit = std::ref(it); // OK?

It turns out that libstdc++ and libc++ do not provide converting assignment operators. In fact, even their converting constructors rely on template type deduction, and so neither of our //OK? lines work on libstdc++ or libc++! On the other hand, MSVC’s standard library makes both of these lines work fine.