[FIXED] How to improve compiler error messages when using C++ std::visit?

Issue

I am using C++17's std::visit() function on a variant with many alternatives, and the error messages produced by the compiler whenever I forget one or more of the alternatives in my visitor are quite difficult to understand.

e.g.

template<class... Ts> struct overloaded : Ts... { using Ts::operator()...; };
template<class... Ts> overloaded(Ts...) -> overloaded<Ts...>;

using Foo = std::variant<A, B, /* ... many more alternatives ... */>;

Foo foo;

std::visit(overloaded{
    [](A const& a) { /* ... */ },
    [](B const& b) { /* ... */ },
    /* ... forgot 1+ alternatives ... */
    }, foo
);

In the above code example, the compiler can produce error messages that are thousands of characters in length, depending on the number of alternatives. Is there a way to improve these error messages so that the compiler will output something like the following instead?

example.cc:8-13: error: Non-exhaustive visitor -- missing alternative of type 'X'

Solution

My first attempt at solving this problem can be found here. After some some googling and lots of trial and error, I've come up with a much better solution, which I've posted here. I'll copy-paste the solution, below, for convenience.

---

Here is a proof of concept.

#include <iostream>
#include <variant>


template <typename> class Test { };

using Foo = std::variant<
    Test<struct A>,
    Test<struct B>,
    Test<struct C>,
    Test<struct D>
    >;

using Bar = std::variant<
    Test<struct E>,
    Test<struct F>,
    Test<struct G>,
    Test<struct H>,
    Test<struct I>,
    Test<struct J>,
    Test<struct K>,
    Test<struct L>
    >;


template <typename T>
struct DefineVirtualFunctor
{
    virtual int operator()(T const&) const = 0;
};

template <template <typename> typename Modifier, typename... Rest>
struct ForEach { };
template <template <typename> typename Modifier, typename T, typename... Rest>
struct ForEach<Modifier, T, Rest...> : Modifier<T>, ForEach<Modifier, Rest...> { };

template <typename Variant>
struct Visitor;
template <typename... Alts>
struct Visitor<std::variant<Alts...>> : ForEach<DefineVirtualFunctor, Alts...> { };


struct FooVisitor final : Visitor<Foo>
{
    int operator()(Test<A> const&) const override { return  0; }
    int operator()(Test<B> const&) const override { return  1; }
    int operator()(Test<C> const&) const override { return  2; }
    int operator()(Test<D> const&) const override { return  3; }
};

struct BarVisitor final : Visitor<Bar>
{
    int operator()(Test<E> const&) const override { return  4; }
    int operator()(Test<F> const&) const override { return  5; }
    int operator()(Test<G> const&) const override { return  6; }
    int operator()(Test<H> const&) const override { return  7; }
    int operator()(Test<I> const&) const override { return  8; }
    int operator()(Test<J> const&) const override { return  9; }
    int operator()(Test<K> const&) const override { return 10; }
    int operator()(Test<L> const&) const override { return 11; }
};


int main(int argc, char const* argv[])
{
    Foo foo;
    Bar bar;
    
    switch (argc) {
    case  0: foo = Foo{ std::in_place_index<0> }; break;
    case  1: foo = Foo{ std::in_place_index<1> }; break;
    case  2: foo = Foo{ std::in_place_index<2> }; break;
    default: foo = Foo{ std::in_place_index<3> }; break;
    }
    switch (argc) {
    case  0: bar = Bar{ std::in_place_index<0> }; break;
    case  1: bar = Bar{ std::in_place_index<1> }; break;
    case  2: bar = Bar{ std::in_place_index<2> }; break;
    case  3: bar = Bar{ std::in_place_index<3> }; break;
    case  4: bar = Bar{ std::in_place_index<4> }; break;
    case  5: bar = Bar{ std::in_place_index<5> }; break;
    case  6: bar = Bar{ std::in_place_index<6> }; break;
    default: bar = Bar{ std::in_place_index<7> }; break;
    }
    
    std::cout << std::visit(FooVisitor{ }, foo) << "\n";
    std::cout << std::visit(BarVisitor{ }, bar) << "\n";

    return 0;
}

As you can see, the Visitor class template accepts a std::variant type as a template parameter, from which it will define an interface that must be implemented in any child classes that inherit from the template class instantiation. If, in a child class, you happen to forget to override one of the pure virtual methods, you will get an error like the following.

$ g++ -std=c++17 -o example example.cc
example.cc: In function ‘int main(int, const char**)’:
example.cc:87:41: error: invalid cast to abstract class type ‘BarVisitor’
   87 |     std::cout << std::visit(BarVisitor{ }, bar) << "\n";
      |                                         ^
example.cc:51:8: note:   because the following virtual functions are pure within ‘BarVisitor’:
   51 | struct BarVisitor final : Visitor<Bar>
      |        ^~~~~~~~~~
example.cc:29:17: note:     ‘int DefineVirtualFunctor<T>::operator()(const T&) const [with T = Test<J>]’
   29 |     virtual int operator()(T const&) const = 0;
      |                 ^~~~~~~~

This is much easier to understand than the error messages that the compiler usually generates when using std::visit().

Answered By - jinscoe123

Answer Checked By - Clifford M. (PHPFixing Volunteer)