Pass class with templated method as template parameter with GCC [duplicate] - c++

Question

This question already has answers here:
C++ template compilation error: expected primary-expression before ‘>’ token [duplicate]
(1 answer)
Where and why do I have to put the "template" and "typename" keywords?
(8 answers)
Closed 4 years ago.
Is it possible to pass object of a class with templated method as a template parameter under GCC? All search I conducted pointed me to questions regarding templated class, which is not applicable in my case.
Sample code:
#include <iostream>
class Foo
{
public:
template<typename T>
void print(const T& arg) const
{
std::cout << arg << std::endl;
}
};
template<typename A>
void print(const A& printer)
{
printer.print<int>(1);
}
int main()
{
Foo foo;
print<Foo>(foo);
return 0;
}
Compiling under MinGW.org GCC-6.3.0-1 I get the following results:
tmpl.cpp: In function 'void print(const A&)':
tmpl.cpp:16:19: error: expected primary-expression before 'int'
printer.print<int>(1);
^~~
tmpl.cpp:16:19: error: expected ';' before 'int'
VisualStudio compiles the same code cleanly, application works as expected. If I make print function not templated, so that is explicitly takes object of Foo it works as expected both compilers.

Answer 1

You need to use template keyword to tell that print is a template.
Inside a template definition, template can be used to declare that a dependent name is a template.
e.g.
template<typename A>
void print(const A& printer)
{
printer.template print<int>(1);
// ~~~~~~~~
}
print is a dependent name, it depends on the template parameter A. Before the template gets instantiated the compiler knows nothing about A; so you need to add the template keyword to tell the compiler that print is a template, then the < after the name print is valid; otherwise < will be treated as the operator< then cause the error.

Answer 2

You need the template keyword in order to specify that print() is a template, like this:
printer.template print<int>(1);