How to mark a constexpr function's parameter unused?

Question

Consider this classic example:

template <typename T, std::size_t N>
constexpr std::size_t arraySize(T (&array)[N]) noexcept { return N; }

Now this works fine, but there is one annoyance, gcc gives a warning:

warning: unused parameter ‘array’ [-Wunused-parameter]

Known solutions:

• Doesn't work: If I add the classic (void)arr; to the function, I get error: body of constexpr function ‘...‘ not a return-statement.
• Unsatisfactory: I can have arraySize(T (&)[N]), but I want to name the argument for two reasons:
  1. It makes compiler error message more understandable.
  2. More subjectively, I think it makes the code clearer, especially to those who don't live and breathe that syntax.
• Not good: In this particular example, I could also return sizeof(array)/sizeof(array[0]);, but this approach is not universal solution, and also I think return N; is much nicer, definitely easier on the eye.
• Good but not always possible: switch to using C++14 and compiler which supports it fully. Then constexpr function body like { (void)array; return N; } is allowed.

How can I get rid of the unused parameter warning nicely, when using C++11?

Answer 1

Try this. I use this method sometimes

template <typename T, std::size_t N>
constexpr std::size_t arraySize(T (& /*array*/ )[N]) noexcept { return N; }