Conditional compilation and non-type template parameters

Question

I am having trouble understanding non-type template arguments and was hoping someone could shed light on this.

#include <iostream>

template<typename T, int a>
void f() {
  if (a == 1) {
    std::cout << "Hello\n";
  } else {
    T("hello");
  }
}

int main() {
  f<int, 1>;
}

When I compile this, I get an error saying:

/tmp/conditional_templates.cc:13:12:   required from here
/tmp/conditional_templates.cc:8:5: error: cast from ‘const char*’ to ‘int’ loses precision [-fpermissive]
     T("hello");
     ^

But, can't the compiler detect that the non-type argument "a" is 1 and hence the else branch won't be taken? Or is that too much to expect? In which case, how do I accomplish something like this?

Answer 1

Try this instead:

#include <iostream>
template<typename T, int a>
struct A {
    void f() {
        T("hello");
    }
};

template<typename T>
struct A<T,1> {
    void f() {
        std::cout << "Hello\n";
    }
};


int main() {
  A<int,1> a;
  a.f();
  A<int,2> b;
  b.f();
}

Now, this uses partial template specialization in order to provide alternative implementations for specific values of the template parameters.

Note that I've used a class, because function templates cannot be partially specialized