Re: Templatized function call interpreted as 'operator<' on gcc
kcsasquatch wrote:
All,
I have the following C++ code that does not compile under gcc 4.2.1 or
4.4.3. It *does* compile on http://comeaucomputing.com/tryitout/
version 4.3.10.1 Beta2.
Is this valid C++ code (as it seems), or is there a real syntax error
that Comeau is not catching?
template<typename T>
class A
{
public:
template<int Flags>
void foo(T *text)
{
return;
}
};
template<typename T>
class B : public A<T>
{
public:
typedef A<const char> base;
template<int Flags>
void foo(T *text)
{
A<T>::foo<Flags>(text); // ERROR
static_cast<A<T>*>(this)->foo<Flags>(text); // OK
base::foo<Flags>(text); // OK
}
};
int main()
{
B<const char> b;
b.foo<1>("bar");
return 0;
}
% g++ main.cpp
main.cpp: In member function ???void B<T>::foo(T*) [with int Flags = 1,
T = const char]???:
main.cpp:32: instantiated from here
main.cpp:23: error: invalid operands of types ???<unresolved overloaded
function type>??? and ???int??? to binary ???operator<???
Only the third one is valid. The second and first one need "->template" and
"::template" respectively. Comeau is not conforming. Note that you can
construct cases that are valid with and without "::template" and produce
different results for each
#include <iostream>
struct A {
template<typename T>
static A f(T) {
return A();
}
template<typename T> operator T() { return T(); }
};
template<typename U>
int g() {
U u;
typedef A (*funcPtrType)(int());
return !(funcPtrType)u.f < int() > (0);
}
int main() {
std::cout << g<A>() << std::endl;
}
Now, that prints 0. But if you insert "template" before "f<int()>" it's
going to putput 1.