Default new alignment

I'm moving some C++17 code to be used in a project that is built with Qt, on Windows, using MinGW 7.3.0, and noticed something weird happening in the 32bit builds:

#include 
#include 
int main() {
  std::cout << __STDCPP_DEFAULT_NEW_ALIGNMENT__ << std::endl;
  std::cout << __BIGGEST_ALIGNMENT__ << std::endl;
  std::cout << sizeof(std::max_align_t) << std::endl;
  std::cout << alignof(std::max_align_t) << std::endl;    
  std::cout << "----------------------------------------" << std::endl;    
  for (int i = 0; i < 8; ++i) {
    const uintptr_t        ptr   = uintptr_t(new char);
    static const uintptr_t first = ptr;
    std::cout << (ptr - first) << " " //
              << (ptr % 16) << " "    //
              << (ptr % 8) << std::endl;
  }    
  return 0;
}

Outputs:

16
16
24
8
----------------------------------------
0 8 0
48 8 0
64 8 0
80 8 0
96 8 0
112 8 0
128 8 0
144 8 0

The same thing happens if I use bigger types / structs instead of char, and also if I call ::operator new(size_t) directly with various sizes, instead of the new operator.

As you can see, STDCPP_DEFAULT_NEW_ALIGNMENT is 16, but I get pointers that are only 8-aligned.

My understanding was that in C++17 operator new(size_t) would always return pointers with alignment of at least STDCPP_DEFAULT_NEW_ALIGNMENT, and the two parameter operator new(size_t, align_val_t) is only required for alignments exceeding STDCPP_DEFAULT_NEW_ALIGNMENT. Here, it seems, operator new(size_t) only respects the lower alignof(std::max_align_t).

Is this expected behavior? What am I getting wrong here? This seems to contradict my reading of: https://en.cppreference.com/w/cpp/memory/new/operator_new