This issue is based on section C++ reference : dependent name - The template disambiguator for dependent name.
I have understood when invoking the template member function in a template class, the keyword template is necessary to make the compiler know the following bracket is used for indicating template argument.just like the used example in this section.
template<typename T>
struct S {
template<typename U> void foo(){}
};
template<typename T>
void bar()
{
S<T> s;
s.foo<T>(); // error: < parsed as less than operator
s.template foo<T>(); // OK
}
However, in the consequent part it describes when a template name appears in a member access expression (after -> or after .), the disambiguator is unnecessary if there is a template with the same name found by ordinary lookup in the context of the expression..
Then it comes with the following code. Comparing with previous example it defines set function whose name exists in standard library as well. At the meanwhile, using std::set is set up to make set template visible in template function. In this condition, even if keyword template is not provided it still works well.
#include <set>
using std::set; // makes 'set' visible to lookup from bar
template<typename T>
struct S {
template<typename U> void set(){}
};
template<typename T>
void bar()
{
S<T> s;
s.set<T>(); // not an error if ::set is visible:
// (and since C++11, this is well-formed)
s.template set<T>(); // works with and without ::set
}
Based on my understanding, I tried my own version
#include <iostream>
template <typename T>
struct S{
template <typename U> void func(){
std::cout << "In S::func\n";
}
};
// In order to make member template function is visible in function test,
// defining a global template function **func** whose name is same with one
// member template function in struct S.
template <typename M>
void func(){
std::cout << "from ordinary func\n";
}
template <typename M>
void test(){
S<M> s;
func<M>(); // test func template function is visible in test function
s.func<M>();
}
int main(){
test<int>();
}
The detail error message is listed as follows
[17:17:50][ryu@C++_test]$ g++ -g typename2.cpp
typename2.cpp:61:7: error: use 'template' keyword to treat 'func' as a dependent
template name
s.func<M>();
^
template
1 error generated.
Any advice is appreciated on how to make my own code works well without keyword template.
