Using `std::experimental::propagate_const` for arrays

Question

The question is simple: why can't I use propagate_const for arrays?

The following line gives errors:

std::experimental::propagate_const<std::unique_ptr<int[]>> ptr = std::make_unique<int[]>(1);

The errors (gcc version 13.0):

/usr/local/include/c++/13.0.0/experimental/propagate_const: In Instanziierung von »struct std::experimental::fundamentals_v2::__propagate_const_conversions<std::unique_ptr<int []> >«:
/usr/local/include/c++/13.0.0/experimental/propagate_const:108:11:   erfordert durch »class std::experimental::fundamentals_v2::propagate_const<std::unique_ptr<int []> >«
../../livews2223/l00/main.cpp:182:64:   von hier erfordert
/usr/local/include/c++/13.0.0/experimental/propagate_const:55:37: Fehler: no match for »operator*« (operand type is »std::unique_ptr<int []>«)
55 |       = remove_reference_t<decltype(*std::declval<_Tp&>())>;
|                                     ^~~~~~~~~~~~~~~~~~~~~

The the goal is to prevent that a const element-function can change array elements like:

struct A {
    void foo() const {
        a[0] = 2; // should not be possible
    }
//    std::experimental::propagate_const<std::unique_ptr<int[]>> a = std::make_unique<int[]>(2);    
    std::unique_ptr<int[]> a = std::make_unique<int[]>(2);    
};

Answer 1

unique_ptr<T[]> is neither a pointer nor a pointer-like type. It's an array-like type. That's why it has operator[] but not operator*. It makes little sense to use *ptr on an array (for language arrays, this accesses the first element, but using ptr[0] makes it much more clear what's going on). So unique_ptr<T[]> provides no such operator.

And therefore, it isn't pointer-like enough for propagate_const to work.