template function within variadic class

Question

Why won't the commented-line in goo compile? Instead I have to resort to defining the global function hoo instead of using the Thing member function foo?

#include <iostream>

template <typename... T>
struct Thing {
    template <typename U> void foo() {std::cout << this << '\n';}
};

template <typename U, typename... T>
void hoo (const Thing<T...>& thing) {std::cout << &thing << '\n';}

template <typename U, typename... T>
void goo (const Thing<T...>& thing) {
//  thing.foo<U>();  // Why won't this line compile?
    hoo<U>(thing);  // Using this instead.
}

int main() {
    Thing<int, double, char> thing;
    goo<short>(thing);
}

What is the change that is needed so that I can use foo() instead?

score 2 · Accepted Answer · answered Dec 19 '14 at 00:40

2

On the line thing.foo<U>() the compiler doesn't have enough information to tell if foo is template or not. Use the template keyword to disambiguate this:

thing.template foo<U>();

answered Dec 19 '14 at 00:40

David G

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Still does not compile (GCC 4.8.1). – prestokeys Dec 19 '14 at 00:47
@prestokeys What's the error? – David G Dec 19 '14 at 00:48
no matching function for call to 'Thing::foo() const'. void Thing::foo() [with U = short int; T = {int, double, char}] no known conversion for implicit 'this' parameter from 'const Thing*' to 'Thing*' – prestokeys Dec 19 '14 at 00:49
@prestokeys `foo()` needs to be `const` since you're calling it on a `const` object. – David G Dec 19 '14 at 00:51
Yes. I realized it. – prestokeys Dec 19 '14 at 00:52

template function within variadic class

1 Answers1