How to detect ubsan presence with g++

Question

I have some valid C++ code that does not compile under ubsan with g++. In a nutshell (https://gcc.godbolt.org/z/9qvz89na8):

struct foo {
  void bar() { }
};

void process(auto f) {
  if constexpr(&decltype(f)::bar);
}

int main() {
  process(foo{});
}

yields

In instantiation of 'void process(auto:1) [with auto:1 = foo]':
error: '(foo::bar != 0)' is not a constant expression

Thus I want to detect if I am building under ubsan, to try to mitigate the issue by making the if constexpr conditionally constexpr.

It's trivial under clang, but with g++ when I try to look at the predefined macros:

echo | g++ -fsanitize=undefined -dM -E -

I cannot see anything that hints that ubsan is enabled. Is there any way to detect this?

Answer 1

Here is an oracle that seems to work.

template<class T>
concept Check = sizeof(int[&T::bar != nullptr]) == sizeof(int[true]);

See: https://gcc.godbolt.org/z/jra1bbY43 for full example.