How does `void_t` work

161
votes

I watched Walter Brown's talk at Cppcon14 about modern template programming (Part I, Part II) where he presented his void_t SFINAE technique.

Example:
Given a simple variable template that evaluates to void if all template arguments are well formed:

template< class ... > using void_t = void;

and the following trait that checks for the existence of a member variable called member:

template< class , class = void >
struct has_member : std::false_type
{ };

// specialized as has_member< T , void > or discarded (sfinae)
template< class T >
struct has_member< T , void_t< decltype( T::member ) > > : std::true_type
{ };

I tried to understand why and how this works. Therefore a tiny example:

class A {
public:
    int member;
};

class B {
};

static_assert( has_member< A >::value , "A" );
static_assert( has_member< B >::value , "B" );

1. has_member< A >

  • has_member< A , void_t< decltype( A::member ) > >
    • A::member exists
    • decltype( A::member ) is well-formed
    • void_t<> is valid and evaluates to void
  • has_member< A , void > and therefore it chooses the specialized template
  • has_member< T , void > and evaluates to true_type

2. has_member< B >

  • has_member< B , void_t< decltype( B::member ) > >
    • B::member does not exist
    • decltype( B::member ) is ill-formed and fails silently (sfinae)
    • has_member< B , expression-sfinae > so this template is discarded
  • compiler finds has_member< B , class = void > with void as default argument
  • has_member< B > evaluates to false_type

http://ideone.com/HCTlBb

Questions:
1. Is my understanding of this correct?
2. Walter Brown states that the default argument has to be the exact same type as the one used in void_t for it to work. Why is that? (I don't see why this types need to match, doesn't just any default type does the job?)

2
c++templatesc++14sfinae
Ad 2) Imagine the static assert was written as: has_member<A,int>::value. Then, the partial specialization that evaluates to has_member<A,void> cannot match. Therefore, it needs to be has_member<A,void>::value, or, with syntactic sugar, a default argument of type void. – dyp
@dyp Thanks, I'll edit that. Mh, I do not see a need in having has_member< T , class = void > defaulting in void yet. Assuming this trait will be used only with 1 template argument at any time, then the default argument could be any type? – nonsensation
Interesting question. – AStopher
Note that In this proposal, open-std.org/jtc1/sc22/wg21/docs/papers/2015/n4436.pdf, Walter changed template <class, class = void> to template <class, class = void_t<>>. So now we are free to do whatever we want with void_t alias template implementation :) – JohnKoch

2 Answers

148
votes

1. Primary Class Template

When you write has_member<A>::value, the compiler looks up the name has_member and finds the primary class template, that is, this declaration:

template< class , class = void >
struct has_member;

(In the OP, that's written as a definition.)

The template argument list <A> is compared to the template parameter list of this primary template. Since the primary template has two parameters, but you only supplied one, the remaining parameter is defaulted to the default template argument: void. It's as if you had written has_member<A, void>::value.

2. Specialized Class Template

Now, the template parameter list is compared against any specializations of the template has_member. Only if no specialization matches, the definition of the primary template is used as a fall-back. So the partial specialization is taken into account:

template< class T >
struct has_member< T , void_t< decltype( T::member ) > > : true_type
{ };

The compiler tries to match the template arguments A, void with the patterns defined in the partial specialization: T and void_t<..> one by one. First, template argument deduction is performed. The partial specialization above is still a template with template-parameters that need to be "filled" by arguments.

The first pattern T, allows the compiler to deduce the template-parameter T. This is a trivial deduction, but consider a pattern like T const&, where we could still deduce T. For the pattern T and the template argument A, we deduce T to be A.

In the second pattern void_t< decltype( T::member ) >, the template-parameter T appears in a context where it cannot be deduced from any template argument.

There are two reasons for this:

  • The expression inside decltype is explicitly excluded from template argument deduction. I guess this is because it can be arbitrarily complex.

  • Even if we used a pattern without decltype like void_t< T >, then the deduction of T happens on the resolved alias template. That is, we resolve the alias template and later try to deduce the type T from the resulting pattern. The resulting pattern, however, is void, which is not dependent on T and therefore does not allow us to find a specific type for T. This is similar to the mathematical problem of trying to invert a constant function (in the mathematical sense of those terms).

Template argument deduction is finished(*), now the deduced template arguments are substituted. This creates a specialization that looks like this:

template<>
struct has_member< A, void_t< decltype( A::member ) > > : true_type
{ };

The type void_t< decltype( A::member ) > can now be evaluated. It is well-formed after substitution, hence, no Substitution Failure occurs. We get:

template<>
struct has_member<A, void> : true_type
{ };

3. Choice

Now, we can compare the template parameter list of this specialization with the template arguments supplied to the original has_member<A>::value. Both types match exactly, so this partial specialization is chosen.

On the other hand, when we define the template as:

template< class , class = int > // <-- int here instead of void
struct has_member : false_type
{ };

template< class T >
struct has_member< T , void_t< decltype( T::member ) > > : true_type
{ };

We end up with the same specialization:

template<>
struct has_member<A, void> : true_type
{ };

but our template argument list for has_member<A>::value now is <A, int>. The arguments do not match the parameters of the specialization, and the primary template is chosen as a fall-back.

(*) The Standard, IMHO confusingly, includes the substitution process and the matching of explicitly specified template arguments in the template argument deduction process. For example (post-N4296) [temp.class.spec.match]/2:

A partial specialization matches a given actual template argument list if the template arguments of the partial specialization can be deduced from the actual template argument list.

But this does not just mean that all template-parameters of the partial specialization have to be deduced; it also means that substitution must succeed and (as it seems?) the template arguments have to match the (substituted) template parameters of the partial specialization. Note that I'm not completely aware of where the Standard specifies the comparison between the substituted argument list and the supplied argument list.

18
votes
// specialized as has_member< T , void > or discarded (sfinae)
template<class T>
struct has_member<T , void_t<decltype(T::member)>> : true_type
{ };

That above specialization exists only when it is well formed, so when decltype( T::member ) is valid and not ambiguous. the specialization is so for has_member<T , void> as state in the comment.

When you write has_member<A>, it is has_member<A, void> because of default template argument.

And we have specialization for has_member<A, void> (so inherit from true_type) but we don't have specialization for has_member<B, void> (so we use the default definition : inherit from false_type)