[LLVMbugs] [Bug 20825] "extern template" seems to invoke explicit instanciation but it should prohibit it.
bugzilla-daemon at llvm.org
bugzilla-daemon at llvm.org
Tue Sep 2 13:17:35 PDT 2014
http://llvm.org/bugs/show_bug.cgi?id=20825
Reid Kleckner <rnk at google.com> changed:
What |Removed |Added
----------------------------------------------------------------------------
Status|NEW |RESOLVED
CC| |rnk at google.com
Resolution|--- |INVALID
--- Comment #1 from Reid Kleckner <rnk at google.com> ---
I think you can forward declare your specialization of the static data member
like so:
// Start .h
template <typename T> class A {
public:
static const char *info;
};
template <> const char *A<int>::info;
template <> const char *A<char>::info;
extern template class A<int>;
extern template class A<char>;
// Start .cpp
#include <iostream>
template <> const char *A<int>::info = "This is int";
template <> const char *A<char>::info = "This is char";
int main(int, char **) {
std::cout << "A<int>:" << A<int>::info << "\n";
std::cout << "A<char>:" << A<char>::info << "\n";
return 0;
}
template class A<int>;
template class A<char>;
----
Otherwise, the compiler can do things like inline the generic, unspecialized
version of an inline function into your program. Consider:
template <typename T>
struct A { static int f() { return sizeof(int); } };
extern template struct A<int>;
int main() {
return A<int>::f();
}
After optimizations, main will return 4, even if some other TU contains:
template <> int A<int>::f() { return 5; };
This specialization can be similarly forward declared as:
template <> int A<int>::f();
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