Why does compiler complain that f() is not visible?

Question

#include <iostream>

using namespace std;

template <size_t N>
typename enable_if<(N > 1), void>::type f(){
    cout << N - 1 << ' ';
    f<N - 1>();
}
template <size_t N>
typename enable_if<N == 1, void> ::type f() {
    cout << 1;
}
int main() {
    f<4>();
}

Compiler complains at line 8:

f< N - 1 >();

Call to function f that is neither visible in the template definition nor found by ADL.

Answer 1

Reverse the order of your function definitions.

#include <iostream>
#include <type_traits>

using namespace std;

template <size_t N>
typename enable_if<N == 1, void> ::type f() {
    cout << 1;
}
template <size_t N>
typename enable_if<(N > 1), void>::type f(){
    cout << N - 1 << ' ';
    f<N - 1>();
}
int main() {
    f<4>();
}

Output:

$ ./a.out
3 2 1 1

Answer 2

Please note that the function is defined below the function call.

You have two possible approaches:

Approach 1:

#include <iostream>
#include <type_traits>

using namespace std;

template <size_t N>
typename enable_if<N == 1, void> ::type f() {
    cout << 1;
}

template <size_t N>
typename enable_if<(N > 1), void>::type f(){
    cout << N - 1 << ' ';
    f<N - 1>();
}

int main() {
    f<4>();
}

Approach 2:

You can forward declare the prototype for the N==1 version function.