The point of uintptr_t is that it's an integer type to which any pointer type can be cast. If you introduce a new class of pointers which are not compatible with uintptr_t, then suddenly you have pointers which are not pointers.
No, uintptr_t is an integer type to which any object pointer type can be converted without loss of information. (Strictly speaking, the guarantee is for conversion to and from void*.) And if an implementation doesn't have a sufficiently wide integer type, it won't define uintptr_t. (Likewise for intptr_t the signed equivalent.)
There's no guarantee that a function pointer type can be converted to uintptr_t without loss of information.
C currently has two kinds of pointer types: object pointer types and function pointer types. "Fat pointers" could be a third. And since a fat pointer would internally be similar to a structure, converting it to or from an integer doesn't make a whole lot of sense. (If you want to examine the representation, you can use memcpy to copy it to an array of unsigned char.)
Surely you're not arguing that a bounded array is in fact a function rather than an object? The distinction between function and object pointers exists for Harvard architecture computers, which sort of exist (old Atmel AVR before they adopted ARM), but are not dominant.
