calling a template parameter constexpr method?

Question

std::array<...>::size() is a non-static constexpr method; being constexpr I can use it as a template parameter:

#include <array>
#include <stdio>

int main(void) {
  std::array<char, 12> a = {{ "Hello world" }};
  std::cout << "size() = " << a.size() << ", data() = \"" << a.data() << "\"" << std::endl;

  std::array<char, a.size()> b = a;
  std::cout << "size() = " << b.size() << ", data() = \"" << b.data() << "\"" << std::endl;
}

However, if the std::array<...> is a template parameter, things become uncertain:

template <typename T>
void copy(T const& a) {
  std::array<char, a.size()> c = a;
  std::cout << "size() = " << c.size() << ", data() = \"" << c.data() << "\"" << std::endl;
}

GCC happily compiles this [1], while CLANG refuses [2] with

<source>:6:20: error: non-type template argument is not a constant expression
  std::array<char, a.size()> c = a;

[1] https://godbolt.org/z/Ru7Y3F
[2] https://godbolt.org/z/LYJcpo

Which one is correct, according to the standard ?

P.S. Yes, I know that I can work around it using std::tuple_size:

template <typename T>
void copy(T const& a) {
  std::array<char, std::tuple_size<T>::value> c = a;
  std::cout << "size() = " << c.size() << ", data() = \"" << c.data() << "\"" << std::endl;
}

score 3 · Answer 1 · answered Aug 21 '18 at 07:30

3

Clang is correct.

You are evaluating an id-expression (a) of reference type that does not has a preceding initialization, which is not allowed in a constant expression.

[expr.const]/2:

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

...

an id-expression that refers to a variable or data member of reference type unless the reference has a preceding initialization and either

it is initialized with a constant expression or

its lifetime began within the evaluation of e;

...

answered Aug 21 '18 at 07:30

xskxzr

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What's a preceding initialization? `a` does get initialized? – Rakete1111 Aug 21 '18 at 16:57
Isn't every reference initialized? In syntax? How else would I initialize a reference? – Rakete1111 Aug 21 '18 at 17:54
@Rakete1111 Sorry I didn't get your point. [This answer](https://stackoverflow.com/a/31720324/5376789) may help. – xskxzr Aug 21 '18 at 18:09
That answer helped a lot! Thanks! So to anyone: When the function is instantiated, `a` doesn't have an initializer (yet). – Rakete1111 Aug 21 '18 at 18:13
@Rakete1111 After reading that answer carefully, I'm also confused with the word "preceding". I'll reply to you if I figure it out. – xskxzr Aug 21 '18 at 18:22
Me too, until I read this comment: https://stackoverflow.com/questions/31714790/c14-initializing-constexpr-variables-from-parameter-values/31720324#comment51384687_31720324 – Rakete1111 Aug 21 '18 at 18:23
Isn't the linked post a dupe? – Passer By Aug 23 '18 at 08:27
@PasserBy I don't know. The question looks not that similar though the answer may be the same. – xskxzr Aug 23 '18 at 08:42
@Rakete1111 I have replied to that comment about my confusion. – xskxzr Aug 23 '18 at 15:08

calling a template parameter constexpr method?

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