I inherit from std::iterator, but compiler does not recognise 'pointer' or 'reference'

Question

I am in the process of making a custom iterator, but I am puzzled on what makes the compiler not recognise the pointer or reference typedefs in the case of a templated iterator implementation.

The following code compiles fine:

#include <iterator>

struct It : std::iterator<std::random_access_iterator_tag, int>
{
    int *v;
    pointer operator-> () { return v; }
    reference operator* () { return *v; }

};

but once I templatize my custom iterator I get errors:

#include <iterator>

template <class T>
struct It2 : std::iterator<std::random_access_iterator_tag, T>
{
    T *v;
    pointer operator-> () { return v; }
    reference operator* () { return *v; }

};

whole code here

error: 'pointer' does not name a type
note: (perhaps 'typename std::iterator<std::random_access_iterator_tag, T>::pointer' was intended)
error: 'reference' does not name a type
...

1> Why can't the compiler see the pointer and reference definitions contained in the std::iterator ?

According to the note, it seems that I shouldn't have bothered using the std::iterator struct, but instead I should have manually copied the typedefs. But that seems too error prone in case the iterator or iterator_traits get an extra typedef in the future.

2. How do you think I should deal with the definition of these traits (pointer, reference etc.) ?

@GrimFandango I'm sorely tempted to correct your misspelling in the title. It's like you can't speak `Merican ;) — Jonathan Mee, May 12 '16 at 15:41

score 6 · Accepted Answer · answered Nov 01 '15 at 15:17

1> Why can't the compiler see the pointer and reference definitions contained in the std::iterator ?

Because the compiler cannot rule out you attempting to provide a specialisation of std::iterator where pointer and reference aren't type definitions, dependent names can never be interpreted as a typedef at template definition time.

2. How do you think I should deal with the definition of these traits (pointer, reference etc.) ?

While explicitly qualifying them with typename every time works, what you can do instead is declare them once as such, and rely on that one declaration in the rest of your It2 definition:

#include <iterator>

template <class T>
struct It2 : std::iterator<std::random_access_iterator_tag, T>
{
    using base = std::iterator<std::random_access_iterator_tag, T>;
    using typename base::pointer;
    using typename base::reference;
    T *v;
    pointer operator-> () { return v; }
    reference operator* () { return *v; }
};

agree. I use this style for all superclasses. It's slightly safer if you make the aliased type `base` private. — Richard Hodges, Nov 01 '15 at 18:41

score 5 · Answer 2 · edited May 23 '17 at 10:34

5

Now that your base class is a dependent type, so are its members pointer and reference.

You need to:

indicate to the compiler that those words are types
indicate to the compiler that those words are member types

You can do it easily:

   typename It2::pointer operator-> () { return v; }
   typename It2::reference operator* () { return *v; }
// ^^^^^^^^^^^^^^

Why? Because C++ is quirky as frak and extremely inexpressive.

edited May 23 '17 at 10:34

Community

1
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answered Nov 01 '15 at 15:06

Lightness Races in Orbit

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Downvotes? [**TRY IT YOURSELF**](http://coliru.stacked-crooked.com/a/f99d21add5dcdbc4) – Lightness Races in Orbit Nov 01 '15 at 15:10
or more accurately, "why? Because how can the compiler otherwise know you *meant* an inherited type? You could have meant some type in the current namespace, a parent namespace or the global namespace." – Richard Hodges Nov 01 '15 at 15:17
2

@RichardHodges: No other language is so weakly constructed so as to require this mess :P – Lightness Races in Orbit Nov 01 '15 at 15:19
@RichardHodges I don't know; how does the compiler know in the first case ? Koenig lookup rules for namespace, but what about global namespace ? – Grim Fandango Nov 01 '15 at 15:25
1

Is there another language that supports the same range of conceptual expressions in a compile-time type-safe way? This is a genuine question. – Richard Hodges Nov 01 '15 at 15:26
@RichardHodges: _Any_ functional programming language, for starters? – Lightness Races in Orbit Nov 01 '15 at 15:30
specifically which one? templates in Haskell are problematic are they not? – Richard Hodges Nov 01 '15 at 16:03
@Richard: Who said anything about templates? But we're going wildly off-topic now. – Lightness Races in Orbit Nov 01 '15 at 16:05
@LightnessRacesinOrbit um, the question was about templates wasn't it? :-) – Richard Hodges Nov 01 '15 at 17:27
1

@RichardHodges: What must be implemented using templates in C++ may be implemented another way in not-C++. You cannot automatically compare like-for-like (which is, indeed, the point). You asked me about another language with the same range of conceptual expressions in a compile-time type-safe way, not one which does so using C++'s notion of templates. – Lightness Races in Orbit Nov 01 '15 at 18:05

score 1 · Answer 3 · answered Nov 01 '15 at 15:09

1

The compiler is very clear about the error:

perhaps typename std::iterator<std::random_access_iterator_tag, T>::pointer was intended

because pointer has become a type-dependent name.

answered Nov 01 '15 at 15:09

edmz

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It's not clear to anyone _why_ "perhaps ... was intended" until explained. – Lightness Races in Orbit Nov 01 '15 at 15:10
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if clang did not provide this alternative, I'd be stuck. Now, I was just puzzled on what is the best way to declare the `operators` – Grim Fandango Nov 01 '15 at 15:20

I inherit from std::iterator, but compiler does not recognise 'pointer' or 'reference'

3 Answers3