I've stumbled over "Why is the template argument deduction not working here?" recently and the answers can be summed up to "It's a nondeduced context".
Specifically, the first one says it's such a thing and then redirects to the standard for "details", while the second one quotes the standard, which is cryptic to say the least.
Can someone please explain to mere mortals, like myself, what a nondeduced context is, when does it occur, and why does it occur?
Deduction refers to the process of determining the type of a template parameter from a given argument. It applies to function templates, auto, and a few other cases (e.g. partial specialization). For example, consider:
template <typename T> void f(std::vector<T>);
Now if you say f(x), where you declared std::vector<int> x;, then T is deduced as int, and you get the specialization f<int>.
In order for deduction to work, the template parameter type that is to be deduced has to appear in a deducible context. In this example, the function parameter of f is such a deducible context. That is, an argument in the function call expression allows us to determine what the template parameter T should be in order for the call expression to be valid.
However, there are also non-deduced contexts, where no deduction is possible. The canonical example is "a template parameter that appears to the left of a :::
template <typename> struct Foo;
template <typename T> void g(typename Foo<T>::type);
In this function template, the T in the function parameter list is in a non-deduced context. Thus you cannot say g(x) and deduce T. The reason for this is that there is no "backwards correspondence" between arbitrary types and members Foo<T>::type. For example, you could have specializations:
template <> struct Foo<int> { using type = double; };
template <> struct Foo<char> { using type = double; };
template <> struct Foo<float> { using type = bool; };
template <> struct Foo<long> { int type = 10; };
template <> struct Foo<unsigned> { };
If you call g(double{}) there are two possible answers for T, and if you call g(int{}) there is no answer. In general, there is no relationship between class template parameters and class members, so you cannot perform any sensible argument deduction.
Occasionally it is useful to inhibit argument deduction explicitly. This is for example the case for std::forward. Another example is when you have conversions from Foo<U> to Foo<T>, say, or other conversions (think std::string and char const *). Now suppose you have a free function:
template <typename T> bool binary_function(Foo<T> lhs, Foo<T> rhs);
If you call binary_function(t, u), then the deduction may be ambiguous and thus fail. But it is reasonable to deduce only one argument and not deduce the other, thus permitting implicit conversions. Now an explicitly non-deduced context is needed, for example like this:
template <typename T>
struct type_identity {
using type = T;
};
template <typename T>
bool binary_function(Foo<T> lhs, typename type_identity<Foo<T>>::type rhs)
{
return binary_function(lhs, rhs);
}
(You may have experienced such deduction problems with something like std::min(1U, 2L).)
Note: std::type_identity is available in the standard library since C++20.
