Re: Templatized function call interpreted as 'operator<' on gcc

"Johannes Schaub (litb)" <>
Tue, 31 Aug 2010 21:24:27 +0200
kcsasquatch wrote:


I have the following C++ code that does not compile under gcc 4.2.1 or
4.4.3. It *does* compile on
version 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

    template<int Flags>
    void foo(T *text)

template<typename T>
class B : public A<T>
    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;<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.

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