constexpr c string concatenation, parameters used in a constexpr context

问题

I was exploring how far I could take the constexpr char const* concatenation from this answer: constexpr to concatenate two or more char strings

I have the following user code that shows exactly what I'm trying to do. It seems that the compiler can't see that the function parameters (a and b) are being passed in as constexpr.

Can anyone see a way to make the two I indicate don't work below, actually work? It would be extremely convenient to be able to combine character arrays through functions like this.

template<typename A, typename B>
constexpr auto
test1(A a, B b)
{
  return concat(a, b);
}

constexpr auto
test2(char const* a, char const* b)
{
  return concat(a, b);
}

int main()
{
  {
    // works
    auto constexpr text = concat("hi", " ", "there!");
    std::cout << text.data();
  }
  {
    // doesn't work
    auto constexpr text = test1("uh", " oh");
    std::cout << text.data();
  }
  {
    // doesn't work
    auto constexpr text = test2("uh", " oh");
    std::cout << text.data();
  }
}

LIVE example

回答1:

concat need const char (&)[N], and in both of your case, type will be const char*, so you might change your function as:

template<typename A, typename B>
constexpr auto
test1(const A& a, const B& b)
{
  return concat(a, b);
}

Demo

来源：https://stackoverflow.com/questions/39199564/constexpr-c-string-concatenation-parameters-used-in-a-constexpr-context