[LLVMbugs] [Bug 18043] constexpr function pointers aren't considered valid template arguments (violates [temp.arg.nontype])

[bugzilla-daemon at llvm.org]

http://llvm.org/bugs/show_bug.cgi?id=18043

M.E. O'Neill <oneill+llvmbugs at cs.hmc.edu> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|RESOLVED                    |REOPENED
         Resolution|INVALID                     |---

--- Comment #2 from M.E. O'Neill <oneill+llvmbugs at cs.hmc.edu> ---
(In reply to comment #1)
> For more background, see:
> 
>   http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#1570
> 
> The code is ill-formed because 'fooAddr' does not refer to a function or an
> array (it refers to a pointer), and thus you cannot omit the '&'. (More
> generally, the intent was that C++11 did not let you do new things here,
> such as pushing the address through a constexpr variable.)

>From my reading, that DR is about *subobjects*.  This is not a subobject, so
the DR does not apply.  

Also, note that your interpretation requires programmers to write this:

template <int x, int y, int z>
void foo()
{
}

template <void (*f)()>
void adapter1()
{
    f();
}

template <void (*f)()>
void adapter2()
{
    f();
}

template <void (*f)()>
void adapter3()
{
    f();
}


void snafu()
{
    adapter1<foo<1234567,31415,271828>>();
    adapter2<foo<1234567,31415,271828>>();
    adapter3<foo<1234567,31415,271828>>();
}

rather than store the desired parameterization of foo in a constexpr constant. 
Also, it's hard to justify why you're reject the above code, but accept this
code

template <int x, int y, int z>
void foo()
{
}

template <void (*f)()>
void adapter1()
{
    f();
}

template <void (*g)()>
void adapter2()
{
    adapter1<g>();
}


void snafu()
{
    adapter2<foo<1234567,31415,271828>>();
}

as g is also a pointer to a function, not the function itself.

So, at the very least you should be applying the rules you believe in
consistently.

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