Is __ptr_t needed?

[Jonathan Wakely <jwakely@redhat.com>]

<bits/cdefs.h> says:

/* This is not a typedef so `const __ptr_t' does the right thing.  */
#define __ptr_t void *

But nothing in glibc headers ever use that macro. I see only three
uses in glibc sources, two of them are comments, and none actually use
const __ptr_t:

sysdeps/mach/hurd/i386/trampoline.c:  uc->uc_stack.ss_sp = (__ptr_t)
sc->sc_uesp;
sysdeps/powerpc/powerpc64/le/power10/memcpy.S:/* __ptr_t [r3] memcpy
(__ptr_t dst [r3], __ptr_t src [r4], size_t len [r5]);
sysdeps/powerpc/powerpc64/multiarch/memcpy-power8-cached.S:/* __ptr_t
[r3] memcpy (__ptr_t dst [r3], __ptr_t src [r4], size_t len [r5]);


Defining this as a macro means that libstdc++ can't use that name,
e.g. for something like:

namespace std { namespace__detail {
  template<typename _Tp>
    using __ptr_t = typename add_pointer<_Tp>::type;
}}

My attempts to squirrel it away in an internal namespace and using a
reserved name fail, because macros laugh at namespaces and crush them
beneath their clumsy feet.

This isn't the end of the world (I can pick another name, or use
#pragma push_macro/pop_macro) but could __ptr_t simply be removed, or
at least changed to a typedef? If any glibc sources do need to use it
for a non-modifiable pointer to void then '__ptr_t const' does the
right thing.

Defining a macro that is barely used seems unnecessary, even if it
does use a reserved name.