Re: Meaning of terms "subexpression" and "constant expression"

From:

James Kanze <james.kanze@gmail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Wed, 2 Feb 2011 15:05:55 CST

Message-ID:

<1129160f-426e-4cd3-873c-481e820cc86a@s29g2000pra.googlegroups.com>

On Feb 1, 9:31 pm, "Johannes Schaub (litb)"
<schaub.johan...@googlemail.com> wrote:

James Kanze wrote:

On Jan 29, 4:39 pm, "Johannes Schaub (litb)"
<schaub.johan...@googlemail.com> wrote:

Nikolay Ivchenkov wrote:

Consider the following example:

#include <iostream>

template <void (*pf)()>
    struct X
{
    template <void (*)()>
        struct Y;
    static void instantiate() { (void)m; }

    typedef Y<&X::instantiate> Inst;

    X() { pf(); }
    static X m;
};

template <void (*pf)()>
    X<pf> X<pf>::m;

void f()
{
    std::cout << "f()\n";
}

The above four lines became:
extern void f();
in the code I talk about below. The rest is unchanged.

int main()
{
sizeof X<&f>();
}

According to N3225 - 3.2/2,

An expression is potentially evaluated unless it is an unevaluated
operand (Clause 5) or a subexpression thereof.

Can the expression f in sizeof X<&f>() be considered as subexpression
of X<&f>()? Is the expression f potentially evaluated and is the
function f odr-used?

I agree, this smells. I can't find what in the spec requires "f" to be
defined either.

The standard clearly says that a definition of f isn't necessary
here. I'm not sure that this is intended, but that's what it
says. At least one compiler accepts the code even if the
definition of f is replaced by a declaration. And off hand,
I can't see why not (even if both g++ and VC++ reject it).

Have you checked whether it correctly links?

Yes.

This code calls f, so I would be rather surprised if it
wouldn't need f to be defined.

Where does it call f? The entire program consists of a single
sizeof. And by definition, anything in the operand of a sizeof
is unevaluated, so nothing in it can be potentially evaluated.

--
James Kanze

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