C++ Nesting SFINAE Template Produces Compilation Error

Question

I'm trying to create an addition operator for a custom template class, where the first argument is allowed to be either an instance of my class or a basic numeric type. My operator has a definition similar to the example code below:

#include <type_traits>

template<typename T>
struct MyTemplateStruct {
    T val;
};

template<typename T, typename U>
struct MyCommonType {
    typedef std::common_type_t<T, U> type;
};
template<typename T, typename U>
using MyCommonTypeT = typename MyCommonType<T, U>::type;

template<typename T, typename U>
MyTemplateStruct<MyCommonTypeT<T, U>> operator +(
    MyTemplateStruct<T> const& a, MyTemplateStruct<U> const& b)
{
    return { a.val + b.val };
}

template<typename T, typename U>
MyTemplateStruct<MyCommonTypeT<T, U>> operator +(
    T const a, MyTemplateStruct<U> const& b)
{
    return { a + b.val };
}

int main()
{
    MyTemplateStruct<double> a{ 0 }, b{ 0 };
    a = a + b;

    return 0;
}

My expectation was that due to SFINAE, the compilation error that results from trying to instantiate the second operator definition with T = MyTemplateStruct<double>, U = double would just exclude that template from the list of potential matches. However, when I compile, I'm getting the following error:

/usr/include/c++/7/type_traits: In substitution of ‘template<class ... _Tp> using common_type_t = typename std::common_type::type [with _Tp = {MyTemplateStruct, double}]’:
main.cpp:18:38: required from ‘struct MyCommonType<MyTemplateStruct, double>’
main.cpp:31:39: required by substitution of ‘template<class T, class U> MyTemplateStruct<typename MyCommonType<T, U>::type> operator+(T, const MyTemplateStruct&) [with T = MyTemplateStruct; U = double]’
main.cpp:40:13: required from here /usr/include/c++/7/type_traits:2484:61: error: no type named ‘type’ in ‘struct std::common_type, double>’

If I directly utilize std::common_type_t in the operator definition, instead of using my wrapper template MyCommonTypeT, then SFINAE works as I expect and there is no error. How can I make the above code compile when I have to wrap the call to std::common_type in another template?

Looking at this answer, it appears I may need to add a template that takes the value true_type/false_type depending on whether my template accepts the provided template parameters. Then, declare my struct to take a defaulted parameter indicating whether the types T/U are supported, and only provide the typedef when they are. Seems kind of kludgy, but it may work. — Jeff G

songyuanyao songyuanyao · Accepted Answer · 2021-09-15T05:53:48

You need to make MyCommonTypeT<T, U> (i.e. MyCommonType<T, U>::type) itself invalid, it's not enough that std::common_type_t<T, U> is invalid when declaring type in MyCommonType. For example you can specialize MyCommonType, when MyTemplateStruct is specified as template argument, type is not declared.

template<typename T, typename U>
struct MyCommonType {
    typedef std::common_type_t<T, U> type;
};
template<typename T, typename U>
struct MyCommonType<MyTemplateStruct<T>, U> {};
template<typename T, typename U>
struct MyCommonType<T, MyTemplateStruct<U>> {};

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C++ Nesting SFINAE Template Produces Compilation Error

1 Answers