Extracts the element of a tuple by type and ambiguity

Question

The article states that for free std::get function overloadings (from 4-6 items) they

Extracts the element of the tuple t whose type is T. Fails to compile if the tuple has more than one element of that type.

Is the last statement a requirements or just a possibility?

I ask the question because std::get< int >(std::make_tuple(1, 2)) compiles fine by clang++ -std=gnu++1z -stdlib=libc++ (live example). Is this tuple's behaviour a libc++ bug or is one conformant to the Standard?

there us something worng with your flags. erase everything and simple write an "Hello world" app with the same flags, the console shows nothing — David Haim, Oct 26 '15 at 09:54
@DavidHaim > Console shows nothing. It is a good result. Lack of the errors during compilation is good information in the case. — Tomilov Anatoliy, Oct 26 '15 at 09:58
@DavidHaim I slightly modified the example to print exit status (zero is success). — Tomilov Anatoliy, Oct 26 '15 at 10:04

Shoe · Answer 1 · 2015-10-26T10:17:09.740

From the standard §20.4.2.6/8, the requirements are:

Requires: The type T occurs exactly once in Types.... Otherwise, the program is ill-formed.

So your program is ill-formed.

The standard even provides an example which is very similar to your situation:

const tuple<int, const int, double, double> t(1, 2, 3.4, 5.6);
const int &i1 = get<int>(t); // OK. Not ambiguous. i1 == 1
const int &i2 = get<const int>(t); // OK. Not ambiguous. i2 == 2
const double &d = get<double>(t); // ERROR. ill-formed

which if you plug in Clang compiles.

Extracts the element of a tuple by type and ambiguity

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