Type conversion at template non-type argument without constexpr

Question

Consider the following code:

struct A {
    constexpr operator int() { return 42; }
};

template 
void foo() {}

void bar(A a) {
    foo(         


        

Answer 1

As @Jarod42 suggests a should be constexpr. This is because templates are deduced at compile time, and so non-type arguments also have to be available at compile time. constexpr is a promise that they will be available at compile time.