Given the following test program:
#include <sstream>
#include <iostream>
class X : private std::ostringstream {
public:
template<typename a> X &operator<<(a arg)
{
#if 0
std::ostringstream &o= *this;
o << arg;
#else
std::ostringstream::operator<<(arg);
#endif
return *this;
}
using std::ostringstream::str;
};
int main(int argc, char **argv)
{
X x;
x << "foo";
std::cerr << x.str() << std::endl;
}
The output that I get is:
$ ./t
0x4014f0
... or some other hexadecimal pointer value.
If I change the "#if 0" to "#if 1" and compile the alternate version, I get the
expected results:
$ ./t
foo
After staring at this for a while, I can't see what the difference is here. It
almost looks like "std::ostringstream::operator<<(arg)" ends up invoking
std::ostringstream::operator<<(void *) even though arg's type is char * (or a
const char *).
--
Summary: Overloaded template losing type in invoking a superclass
method?
Product: gcc
Version: 4.1.2
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
AssignedTo: unassigned at gcc dot gnu dot org
ReportedBy: mrsam at courier-mta dot com
GCC build triplet: x86_64-redhat-linux
GCC host triplet: x86_64-redhat-linux
GCC target triplet: x86_64-redhat-linux
http://gcc.gnu.org/bugzilla/show_bug.cgi?id=34696
