CODEWITHSUNDEEP
c++templatesc++11constexprone-definition-rule
marlam
marlam

Reputation: 610

Assign static constexpr class member to runtime variable

I know there are a lot of similar questions, but somehow different questions. It is about the following situation:

#include <iostream>
#include <array>

template<typename T> class MyClass
{
public:
    static constexpr std::array<T,4> ARRAY {{4, 3, 1, 5}};
};

int main()
{
    constexpr std::array<int, 4> my_array(MyClass<int>::ARRAY); // works fine -> can use the ARRAY to initialize constexpr std::array

    constexpr int VALUE = 5*MyClass<int>::ARRAY[0]; // works also fine

    int value;
    value = my_array[0]; // can assign from constexpr
    value = MyClass<int>::ARRAY[0]; // undefined reference to `MyClass<int>::ARRAY

    std::cout << VALUE << std::endl;
    std::cout << value << std::endl;

    return 0;
}

As far as I understand constexpr is for compile-time constants. So the compiler can do already some calculation, for example to calculate the VALUE. Also I can obviously define a constexpr std::array<,>, from which I can assign the values to runtime variables. I would expect the compiler to set already value = 4 into the executable program, to avoid a loading operation. However, I cannot assign directly from the static member, getting the error

undefined reference to `MyClass<int>::ARRAY'
clang-3.7: error: linker command failed with exit code 1

which makes no sense to me, because it can be done with an intermediate step of another constexpr variable.

So my question is: Why can a static constexpr member of a class not be assigned to a runtime variable?

Note: In my MWE the class is a template class, which does not affect the error. However, I was originally interested in this particular case, which I expect to be more general as for a non-template class.

(Compiler is clang++ or g++ with -std=c++11 - they give the same error)

Edit: @Bryan Chen: Forgot the output lines. Are added now.

Upvotes: 8

Views: 1490

Answers (2)

M.M
M.M

Reputation: 141534

The undefined reference is a linker error. The rule is that if a variable is odr-used then it must have a definition. This applies even for constexpr variables.

Like most ODR rules, violating it is undefined behaviour with no diagnostic required (which could explain why you saw no diagnostic for some of your uses of the value).

To fix the error, add a definition outside the class:

template<typename T> constexpr std::array<T,4> MyClass<T>::ARRAY;

Since it is a template you can actually put this in the header, as opposed to the usual case where the definition goes in exactly one .cpp file.

The main issue here is whether ARRAY[0] counts as odr-use. According to this detailed post, in C++11 and C++14, indexing an array does count as odr-use , but this was changed by DR 1926 filed against C++14 to not be odr-use.

However, that is talking about builtin arrays. IDK whether the same rationale applies to std::array, I find the text of [basic.def.odr]/3 hard to understand. According to the informal definition on cppreference, std::array::operator[] would cause odr-use of the array because its return value binds a reference to the array.

Upvotes: 7

Richard Hodges
Richard Hodges

Reputation: 69854

For this reason I always return constexpr objects from a constexpr function.

Modified code below. Note that due to a c++14 deficiency in std::array<> you must return a const std::array in order to allow operator[] to work.

#include <iostream>

#include <iostream>
#include <array>

template<typename T> class MyClass
{
public:
    static constexpr const std::array<T,4> ARRAY() { return {4, 3, 1, 5}; };
};

int main()
{
    constexpr std::array<int, 4> my_array(MyClass<int>::ARRAY()); // works fine -> can use the ARRAY to initialize constexpr std::array

    constexpr int VALUE = 5 * MyClass<int>::ARRAY()[0]; // works also fine

    int value;
    value = my_array[0]; // can assign from constexpr
    value = MyClass<int>::ARRAY()[0]; // undefined reference to `MyClass<int>::ARRAY

    std::cout << VALUE << std::endl;
    std::cout << value << std::endl;

    return 0;
}

expected results:

20
4

Upvotes: 4

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