Concat two `const char` string literals

Question

Is it possible to concat two string literals using a constexpr? Or put differently, can one eliminate macros in code like:

#define nl(str) str "\n"

int main()
{
  std::cout <<
      nl("usage: foo")
      nl("print a message")
      ;

  return 0;
}

Update: There is nothing wrong with using "\n", however I would like to know whether one can use constexpr to replace those type of macros.

Answer 1

1. Yes, it is entirely possible to create compile-time constant strings, and manipulate them with constexpr functions and even operators. However,

2. The compiler is not required to perform constant initialization of any object other than static- and thread-duration objects. In particular, temporary objects (which are not variables, and have something less than automatic storage duration) are not required to be constant initialized, and as far as I know no compiler does that for arrays. See 3.6.2/2-3, which define constant initialization, and 6.7.4 for some more wording with respect to block-level static duration variables. Neither of these apply to temporaries, whose lifetime is defined in 12.2/3 and following.

So you could achieve the desired compile-time concatenation with:

static const auto conc = <some clever constexpr thingy>;
std::cout << conc;

but you can't make it work with:

std::cout <<  <some clever constexpr thingy>;

Update:

But you can make it work with:

std::cout << *[]()-> const {
             static constexpr auto s = /* constexpr call */;
             return &s;}()
          << " some more text";

But the boilerplate punctuation is way too ugly to make it any more than an interesting little hack.

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(Disclaimer: IANALL, although sometimes I like to play one on the internet. So there might be some dusty corners of the standard which contradicts the above.)

(Despite the disclaimer, and pushed by @DyP, I added some more language-lawyerly citations.)