Re: , definition of "used" for templates

From:
=?UTF-8?B?RGFuaWVsIEtyw7xnbGVy?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Fri, 8 Oct 2010 15:49:16 CST
Message-ID:
<i8ns7h$5ll$1@news.eternal-september.org>
Am 08.10.2010 15:48, schrieb Johannes Schaub (litb):
[..]

More interestingly, what about this one?

struct Printer { Printer() { std::cout<< "I do print!"<<
std::endl; } };

template<typename T> struct A { static Printer printer; };

template<typename T> Printer A<T>::printer;

template<typename T> void f() { &A<void>::printer; }

int main() { }

If that line "uses" the printer because it appears in a potentially
evaluated expression, it shall print "I do print!". I tested this on
both GCC and Clang, both of which do *not* print "I do print!".

So it appears GCC and Clang disagree with your reading of the ODR.


I don't think that they disagree. The wording of the combination
of 3.2 and 14.7.1 is sufficiently diffuse to give a lot of
wiggle room for implementations. For a user it means that the
variable is potentially used, for an implementation the wording
allows to not instantiate A<void> *and* not to use A<void>::printer
(Note my emphasize that both decisions are independent ones).

The wording basically allows an implementation to "rewrite" f()
as follows:

void __invoker() { // Existing as a single definition
  &A<void>::printer;
}

template<typename T>
void f() {
  __invoker();
}

and to use this as an opportunity to prevent multiple instantiations
of &A<void>::printer when f() *would* be instantiated multiple
times in another program - it's just so that most (all?)
implementations don't do that this way.

I certainly agree that the current situation is quite pessimistic
for a user and especially the combination of ODR and template
instantiation rooms should be more clearly defined.

HTH & Greetings from Bremen,

Daniel Kr??gler

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