c++ - How do I call a templated member function without knowing whether or not the class is a template? -

suppose have following macro:

#define call_foo(x) x.foo<int>(); 

this macro call templated foo function template argument int, , compile long x not template type. if is, need define second variant of macro handle template types:

#define call_foo_templated(x) x.template foo<int>(); 

how can create macro call template member function foo regardless of whether or not x dependent name?

the answer you're looking for, in [temp.names], template required when:

when name of member template specialization appears after . or -> in postfix-expression or after nested-name-specifier in qualified-id, , object expression of postfix-expression type-dependent or nested-name-specifier in qualified-id refers dependent type, name not member of current instantiation (14.6.2.1), member template name must prefixed keyword template.

but while there restrictions on name referred to:

a name prefixed keyword template shall template-id or name shall refer class template.

there not restrictions on "dependence-ness" of it:

[ note: case typename prefix, template prefix allowed in cases not strictly necessary; i.e., when nested-name-specifier or expression on left of -> or . not dependent on template-parameter, or use not appear in scope of template. —end note]

so following compile fine:

struct x {     template <typename >     void foo() { } };  int main() {     x{}.template foo<int >(); // template keyword not strictly necessary } 

the real answer though is: don't use macro such thing. write out function call. thank later. added bonus @ call site, know whether or not template keyword required, can use precisely necessary.