if constexpr instead of tag dispatch

Question

I want to use if constexpr instead of tag dispatching, but I am not sure how to use it. Example code below.

template<typename T>
struct MyTag
{
   static const int Supported = 0;
};

template<>
struct MyTag<std::uint64_t>
{
  static const int Supported = 1;
};

template<>
struct MyTag<std::uint32_t>
{
  static const int Supported = 1;
};

class MyTest
{
public:
   template<typename T>
   void do_something(T value)
   {
      // instead of doing this
      bool supported = MyTag<T>::Supported;

      // I want to do something like this
      if constexpr (T == std::uint64_t)
          supported = true;
   }
};

You can't "call" `operator==` on types; that is meaningless. You could do something similar with a library like boost hana: `if constexpr (hana::type_c == hana::type_c)`. You can also just use `std::is_same`: `if constexpr (std::is_same_v)` — Justin, Sep 12 '17 at 06:49
@Rakete1111 Yeah I know that I cannot compare them like that, that is what the question is all about, how I should do it.... and why the downvote? — 0xBADF00, Sep 12 '17 at 06:50
@Justin That looks like a decent answer to me, care to post it as such? — , Sep 12 '17 at 07:01

Richard Hodges · Accepted Answer · 2017-09-12T07:28:06.563

One way is to define a constexpr predicate which checks the type of its argument, then constexpr switch on the result of that predicate.

I think this way is nice because it separates the functional logic from the precondition logic.

#include <iostream>
#include <cstddef>
#include <type_traits>

class MyTest
{
public:
    template<typename T>
    void do_something(T value)
    {
        // define our predicate
        // lambdas are constexpr-if-possible in c++17
        constexpr auto is_supported = [](auto&& x) {
            if constexpr (std::is_same<std::decay_t<decltype(x)>, std::uint64_t>())
                return true;
            else
                return false;
        };

        // use the result of the predicate        
        if constexpr (is_supported(value))
        {
            std::cout << "supported\n";
        }
        else 
        {
            std::cout << "not supported\n";
        }
    }
};

int main()
{
    auto t = MyTest();

    t.do_something(int(0));
    t.do_something(std::uint64_t(0));
    t.do_something(double(0));
    t.do_something(static_cast<unsigned long>(0));  // be careful with std::uint_xx aliases

}

example results:

not supported
supported
not supported
supported

Another way to express this might be:

class MyTest
{
public:

    template<class T>
    static constexpr bool something_possible(T&&)
    {
        return std::is_same<std::decay_t<T>, std::uint64_t>();
    }

    template<typename T>
    void do_something(T value)
    {
        // switch behaviour on result of constexpr predicate    
        if constexpr (something_possible(value))
        {
            std::cout << "supported\n";
        }
        else 
        {
            std::cout << "not supported\n";
        }
    }
};

thanks very nice answer, and the std::decay is needed if T is a pointer or reference, right? — 0xBADF00, Sep 12 '17 at 08:11
@0xBADF00 the decay strips off const, reference, etc so you can compare the underlying types. — Richard Hodges, Sep 12 '17 at 08:14
@RichardHodges this has some sense for generic lambdas but for "ordinary" functions one can use template values matching rules to stop compilation for illegal types. — , May 18 '18 at 09:25
@RichardHodges or even just declare method with a required type. Ditto, an functional/lambdas example might be better ... — , May 18 '18 at 09:33

score 0 · Answer 2 · answered May 18 '18 at 11:14

Usually runtime interrogation of types has sense in functional programing with generic lambdas (with generic arguments too). Otherwise simple answer might be: just declare using 'required' types or use type traits, etc ... Back to the subject of generic lambdas.

/// <summary>
/// c++ 17 generic lambdas have issues
/// with required types of auto arguments
/// in c++20 this will be fixed with new
/// lambda arguments template declaration syntax
/// until then ...
/// </summary>
namespace required_types
{
    template<typename RQ>
    inline auto  is_required_type = [](const auto & v_ = 0) constexpr -> bool
    {
        using T = std::decay_t< decltype(v_) >;
        return std::is_same<T, RQ>();
    };


    inline auto is_uint64 = [] ( const auto & v_ = 0 ) constexpr -> bool
    {
        return is_required_type<std::uint64_t>(v_);
    };

} // required_types

namespace {

    using namespace required_types;

    inline auto tv = [](const char prompt[] = "", const auto & value) {
        std::cout << prompt << "\ntype:\t" << typeid(decltype(value)).name() << "\nvalue:\t" << value;
    };


    inline auto make_double_value = [](auto value)
    {
        if constexpr (is_uint64(value)) {
            tv("\n\nDoubling required type (std::uint_64):", value);
            return value + value;
        }

        tv("\n\nWill try to double 'illegal' type", value);
        return value + value;
    };

}

some usage

        // call with 'legal' aka required type
    std::uint64_t u42 = 42u;
    auto double_value_2 = make_double_value(u42);
    tv("\nResult:", double_value_2);

    // call with some 'illegal' types also works
    auto double_value = make_double_value(42u);
    tv("\nResult:", double_value);

    std::string one{"--ONE--"};
    auto double_value_3 = make_double_value(one);
    tv("\nResult:", double_value_3 );

Of course if one hotly disagrees with my intro one can still use my "required_types":

template<typename T>
void some_proc ( const T && val_ ) {
    using namespace required_types;
      if constexpr ( is_required_type<std::uint64_t>(val_) ) {
            do_something_with_uint64 (val_) ;
      } 
}

Instead of above I would much rather use std::enable_if, somewhere along this answer.

But (as mentioned) for solving few generic lambdas issues in C++17 I would (boldly) use my namespace required_types, with some extensions.

if constexpr instead of tag dispatch

2 Answers2

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