Does different partial specialization in different compiler unit defined behavior in C++?

Question

For example, suppose I have a template class foo in foo.h like below:

template <typename T>
struct foo {
    constexpr static auto value = false;
};

in first.cpp file, I use the foo like below:

#include "foo.h"
auto dummy1() {
    return foo<int>::value;
}

in second.cpp file, I write partial specialization for foo and use foo like below:

#include "foo.h"
template <std::integral T>
struct foo<T> {
    constexpr static auto value = true;
};

auto dummy2() {
    return foo<int>::value;
}

It's such program a well defined behavior in c++? What if dummy2 use unsigned int like below:

auto dummy2() {
    return foo<unsigned int>::value;
}

Answer 1

That's explicit speciliaztion, not partial specialization, and yes it makes the program IFNDR (ill-formed, no diagnostic required) which effectively means the same as undefined behavior for all inputs.

See [temp.expl.spec]/7 which requires the explicit specialization to be declared before first use that would cause implicit instantiation of the specialization in every translation unit where such use is present.

---

After the update of the question:

Now it is indeed a partial specialization, but basically the same holds. See [temp.class.spec.general]/1 for the equivalent sentence regarding partial specializations.

So you need to at least add a declaration of the partial specialization to the header:

template <typename T>
struct foo {
    constexpr static auto value = false;
};

template <std::integral T>
struct foo<T>;

Although then it won't be enough to resolve foo<int>::value, so you are back to needing to put the whole definition of the partial specialization into the header.