How to ensure constexpr function never called at runtime?

In C++20 you can just replace constexpr by consteval to enforce a function to be always evaluated at compile time.

Example:

          int    rt_function(int v){ return v; }
constexpr int rt_ct_function(int v){ return v; }
consteval int    ct_function(int v){ return v; }

int main(){
    constexpr int ct_value = 1; // compile value
    int           rt_value = 2; // runtime value

    int a = rt_function(ct_value);
    int b = rt_ct_function(ct_value);
    int c = ct_function(ct_value);

    int d = rt_function(rt_value);
    int e = rt_ct_function(rt_value);
    int f = ct_function(rt_value); // ERROR: runtime value

    constexpr int g = rt_function(ct_value); // ERROR: runtime function
    constexpr int h = rt_ct_function(ct_value);
    constexpr int i = ct_function(ct_value);
}

---

Pre C++20 workaround

You can enforce the use of it in a constant expression:

#include<utility>

template<typename T, T V>
constexpr auto ct() { return V; }

template<typename T>
constexpr auto func() {
    return ct<decltype(std::declval<T>().value()), T{}.value()>();
}

template<typename T>
struct S {
    constexpr S() {}
    constexpr T value() { return T{}; }
};

template<typename T>
struct U {
    U() {}
    T value() { return T{}; }
};

int main() {
    func<S<int>>();
    // won't work
    //func<U<int>>();
}

By using the result of the function as a template argument, you got an error if it can’t be solved at compile-time.