Re: Is finding definitions of explicitly specialized template
functions compiler-defined?
On Apr 2, 6:03 pm, "jason.cipri...@gmail.com"
<jason.cipri...@gmail.com> wrote:
Here is an example with 3 files, containing a template
structure and also a template function. The header A.h
declares a template structure A with a default (i.e. for any
template parameter), inlined constructor implementation, and
also declares a template function f. The file main.cpp
contains some test code and also the default implementation
for f(). The file spec.cpp contains some explicitly
specialized stuff:
==== A.h ====
#include <iostream>
using std::cout;
using std::endl;
template <int> struct A {
A () { cout << "default" << endl; }
};
template <int> void f ();
==== main.cpp ====
#include "A.h"
template <int> void f () { cout << "default" << endl; }
int main () {
A<0> x;
A<1> y;
f<0>();
f<1>();
return 0;
}
==== spec.cpp ====
#include "A.h"
template<> A<1>::A () { cout << "A 1!" << endl; }
template<> void f<1> () { cout << "f 1!" << endl; }
==== END EXAMPLE ====
When using GCC, if I only compile main.cpp:
g++ main.cpp
Running the program produces the output:
default
default
default
default
Which makes sense. The default A::A() and f() implementations
print the word "default". If I simply add spec.cpp:
g++ main.cpp spec.cpp
The linker seems to know that since explicit specializations
of certain functions are present in other object files, it
should use them instead of the default implementations, and
the output is:
default
A 1!
default
f 1!
I have not tried this with any other compilers besides GCC
4.1.2. My question is: Is the behavior of the linker here
specific to GCC's linker? Or is it always guaranteed that if
one object file contains definitions for explicit
specializations of template functions, and that object file is
linked, then they will be used everywhere? The biggest reason
that I'm unsure is the linker has to do a small amount of
magic to make this work; falling back on the default
implementation if no explicit specializations were found, so
I'm wondering if that magic is defined by C++ or is a GCC
implementation detail.
=A714.7.3/6: "If a template, a member template or the member of a
class template is explicitly specialized then that
specialization shall be declared before the first use of that
specialization that would cause an implicit instantiation to
take place, in every translation unit in which such a use
occurs; no diagnostic is required." In other words, undefined
behavior.
FWIW: I think that in main.cpp, g++ instantiates your template,
affecting the instantiation to a weak symbol definition. In
spec.cpp, the specialization generates a normal symbol
definition. The rules of the linker are to first resolve using
a normal symbol definition, if one exists (and more than one is
an error), then try using weak definitions (choosing one more or
less at random if more than one occurs). I think that this is a
relatively common implementation, so there is a fairly good
chance your code will actually work. It's still undefined
behavior, however, and I don't think even g++ actually
guarantees it. In other words, referring to your subject line,
it's not even compiler-defined; it just happens to work that
way, as a side effect of certain implementation choices, which
might change in the next version of the compiler.
The usual solution when explicit specializations are present is
to declare them, either in the header with the template, or in
the case of a specialization on a type argument, in the header
which defines the type.
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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