Detecting in C++ whether a type is defined, part 2: Giving it a special name

Raymond Chen

Raymond

Warning to those who got here via a search engine: Don’t use this version. Keep reading to the end of the series.

Last time, we detected whether a type was defined by setting up the unqualified name search order so that the name search would find the type if it were defined, or a fallback type if not. One problem with that technique was that the search had to be done from the specially-constructed namespace.

So let’s fix that, and build on the result.

// awesome.h
namespace awesome
{
  // might or might not contain
  struct special { ... };
}

// your code
namespace detect::impl
{
  struct not_implemented {};
  using special = not_implemented;
}

namespace awesome::detect
{
  using namespace ::detect::impl;
  using special_maybe = special;
}

This time, I introduce a new type special_maybe. I did it inside the awesome::detect namespace, so the name special on the right hand side undergoes unqualified name lookup, like we described last time, and it will pick either the defined type ::awesome::special or the fallback type ::detect::impl::special. You can then use some new helpers:

namespace detect
{
  template<typename T>
  constexpr bool is_defined_v =
    !std::is_same_v<T, impl::not_implemented>;
}

void foo()
{
 if constexpr (detect::is_defined_v
                <awesome::detect::special_maybe>) {
   // do something now that we know "special" exists.
 }
}

This looks like it would work great, but it doesn’t. Because inside the “do something now that we know special exists”, you probably want to use special. But you can’t, because special might not exist!

While it’s true that if constexpr tells the compiler to discard the not-taken branch, the code in the not-taken branch must still be valid. If special is not defined, then the body of the if constexpr will contain references to the nonexistent entity special, so it will not compile. You could try using special_maybe, but that’s just a dummy type, and it won’t have the methods you want to call.

So we have to play a trick: Use the type without saying the type!

template<typename T, typename TLambda>
void call_me_maybe(TLambda&& lambda)
{
  if constexpr (is_defined_v<T>) {
    lambda(static_cast<T*>(nullptr));
  }
}

This helper function doesn’t look all that useful. I mean, if the type exists, then we call the lambda. That just puts us back where we started, doesn’t it? I mean, the lambda will need to use the type, which it can’t do if the type doesn’t exist.

Not quite. Because it lets us do this:

void foo(Source const& source)
{
  call_me_maybe<awesome::detect::special_maybe>(
    [&](auto* p)
    {
      using T = std::decay_t<decltype(*p)>;
      T::static_method();
      auto s = source.try_get<T>();
      if (s) s->something();
    });
}

What’s going on?

The way C++ lambdas work is that a lambda becomes an anonymous type with an operator() method. For your typical lambda, the operator() is a const-qualified method whose prototype matches that of the lambda:

auto lambda1 = [](int v) -> void { ... };

// becomes

struct anonymous1
{
 auto operator()(int v) -> void const { ... };
};
auto lambda1 = anonymous1();

However, if the parameter list uses auto, then the operator() itself becomes a template function:

auto lambda2 = [](auto v) -> void { ... };

// becomes

struct anonymous2
{
 template<typename T>
 auto operator()(T v) -> void const { ... };
};
auto lambda2 = anonymous2();

Next, we take advantage of the fact that in a template function, entities that are dependent upon the template parameter are not resolved until the template is instantiated. In this case, the template function is the operator() of the lambda.

This means that our lambda body can do things that depend on the type of p, and the compiler can’t validate that those things are meaningful until the template is instantiated, because it isn’t until that time that the templated operator() is instantiated and the compiler knows what type it needs to use.

In other words, we use the incoming parameter merely for its type information. We extract the type of the pointed-to object and call it T. Then whenever we would normally say special, we just say T

And then we realize that we don’t have to call it T. We can call it… special!

void foo(Source const& source)
{
  call_me_maybe<awesome::detect::special_maybe>(
    [&](auto* p)
    {
      using special = std::decay_t<decltype(*p)>;
      special::static_method();
      auto s = source.try_get<special>();
      if (s) s->something();
    });
}

With this little change, the code inside the lambda looks pretty much like the code you would have written all along, with the bonus feature that it’s legal code even if special doesn’t exist!

We’re getting closer. Next time, we’ll get rid of all this maybe nonsense.

¹ C++20 makes this a little easier by letting us get the type directly, rather than having to extract it from the parameter.

void foo(Source const& source)
{
  call_me_maybe<awesome::detect::special_maybe>(
    [&]<typename T>(T*)
    {
      T::static_method();
      auto s = source.try_get<T>();
      if (s) s->something();
    });
}

Or, using the second trick:

void foo(Source const& source)
{
  call_me_maybe<awesome::detect::special_maybe>(
    [&]<typename special>(special*)
    {
      special::static_method();
      auto s = source.try_get<special>();
      if (s) s->something();
    });
}

 

Raymond Chen
Raymond Chen

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4 comments

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  • Avatar
    Keith Patrick

    I remember after trying to go back to (Managed) C++ after years of Java and C#, and I did not make it very far.  I found a lot of the syntax cluttered and nearly unreadable (scopes, I’m looking at::you), which was made worse by the Managed part.  I could have adjusted to all of that, but where I gave up was missing operators like ‘is/as’ and all the reflection stuff.  It makes me wonder how I got around that when I was doing C++ full-time.

    • Avatar
      Me Gusta

      As someone who is still doing C++ full time, it makes me wonder why you think you need is/as.
      If you really need the dynamic type then you can easily reproduce these operators using typeid or dynamic_cast, but these operators are slow. This results in C++ code not using unknown objects anywhere near as much so there is no need for is/as or reflection.
      You seemingly have gotten into the C# mindset for programming, and this isn’t the same as the C++ mindset for programming.

      • Avatar
        Keith Patrick

        Yep, your analysis is spot-on.  I don’t even know how long it would take me to return to the C++ mindset (I really miss const parameters and the option to use stack-allocated memory)…I ponder it every now and then when I get emails about C++ jobs.  It is crazy how utterly dependent on type information I’ve become once I got my hands on it.