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Invoking `constexpr` member function through reference - clang vs gcc


By : Iwan Riwayanto
Date : July 31 2020, 04:00 PM
fixed the issue. Will look into that further Consider the following example (snippet (0)): ,
Is g++ 9.x correct in accepting snippet (0)?
code :


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C++ constexpr auto member function. Clang issue?


By : user3600825
Date : March 29 2020, 07:55 AM
it should still fix some issue , If you consider this statement in [dcl.constexpr]/7:

Constexpr member function returning a template parameter not considered constexpr when accessed through a reference


By : treg11
Date : March 29 2020, 07:55 AM
I wish this helpful for you Something passed by reference cannot be constexpr. And, for a function to actually be used as constexpr all of the arguments and objects used within the function must also be constexpr. Now, since a reference (even of a template object) can never be constexpr, this means that the better choice is for the compiler to throw an error and let the developer know, rather than the alternative which is to generate the non-constexpr version of the function.
This is because reference means it must have a runtime memory location (see C++11 rvalue and lvalue, and rvalue references). If something qualifies as constexpr it means it is only known and used at compile time, and the result is baked into the resulting executable the same way as hardcoded numbers or strings.

constexpr non-static member function with non-constexpr constructor (gcc,clang differ)


By : Vanthar
Date : March 29 2020, 07:55 AM
this one helps. It's a core defect that was fixed for C++14
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#1684

Clang claims constexpr member of generic lambda argument is not constexpr


By : user2964701
Date : March 29 2020, 07:55 AM
wish helps you Since this_r is a static member you can always access it without dereferencing (non constexpr) reference to object instance to make clang or other compilers happy:
code :
int f(std::variant<S<0>, S<1>, S<2> > v) {
    return std::visit([](auto const& arg) {
        if constexpr(::std::remove_reference_t<decltype(arg)>::this_r == 0) { return 42; }
        else { return ::std::remove_reference_t<decltype(arg)>::this_r; }
    }, v);
}

Constexpr member function on l-value ref object: Clang and gcc disagree


By : Vishal Verma
Date : March 29 2020, 07:55 AM
I wish this help you As specified by the definition of core constant exprssions:
code :
using A = std::array<int, 10>;

constexpr A arr = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
constexpr const A& arr_ref = arr;   // OK: arr_ref is initialized with a constant expr

static_assert(arr.size() > 0, "");
static_assert(arr_ref.size() > 0, "");
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