Is use in an unused default member initializer still an odr-use?

Question

Is use in a default member initializer still an odr-use, even if the default member initializer is not used by any constructor?

For example, is this program ill-formed because g<A> is odr-used and therefore its definition implicitly instantiated?

template<typename T>
void g() { sizeof(T); }

struct A;

struct B {
    B() : i{} {};

    int i = (&g<A>, 0);
};

int main() { }

MSVC thinks no. Clang, GCC and ICC think yes. https://godbolt.org/z/zrr9oEdfe

`g` is [potentially-evaluated](https://timsong-cpp.github.io/cppwp/basic.def.odr#3), and it does not fall into any of the exceptions listed in the [definition of odr-use](https://timsong-cpp.github.io/cppwp/basic.def.odr#5). Therefore it is odr-used. All compilers are behaving correctly, even MSVC, because odr violations are classified as "[no diagnostic required](https://timsong-cpp.github.io/cppwp/basic.def.odr#11)". — Raymond Chen, Mar 12 '22 at 19:38
@RaymondChen There is a reachable definition for `g` which I think should be implicitly instantiated if the function is odr-used. That instantiation would be ill-formed, so I thought a diagnostic should be given. — user17732522, Mar 12 '22 at 20:23

score 8 · Accepted Answer · answered Mar 13 '22 at 00:25

As stated in the comments, g<A> is odr-used. However, there is a definition for it available, so there is no non-diagnosable violation here; MSVC is wrong to accept it. (This is true even without the constructor declaration; the implicitly declared B::B() is never defined, but the default member initializer is still an odr-use like it is here.)

Is use in an unused default member initializer still an odr-use?

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