C++ constexpr in place aligned storage construction

Question

I'm trying to make an aligned variant type that uses std::aligned_storage to hold the data. Is there a way to construct an object in place in a constexpr way? I read you can't do constexpr placement new.

#include <iostream>
#include <string>


struct foo
{
    foo(std::string a, float b)
    : bar1(a), bar2(b)
    {}

    std::string bar1;
    float bar2;
};


struct aligned_foo
{
    template<typename... Args>
    aligned_foo(Args&&... args) 
    {
        //How to constexpr construct foo?
        data_ptr = ::new((void*)::std::addressof(storage)) foo(std::forward<Args>(args)...);
    }

    std::aligned_storage<sizeof(foo)> storage;
    foo* data_ptr;
};


int main()
{
    aligned_foo("Hello", 0.5);
}

`std::string` doesn't have a `constexpr` constructor, so unlikely. — Bo Persson, Oct 28 '17 at 11:15
**Short answer:** No. **Long answer:** One of the possible reasons: The lifetime of an object of a class type begins when its constructor has completed. For `constexpr` contexts, the only sane way to manage such lifetimes is with objects of *automatic storage duration* or *static storage duration* - because the compiler can reason about these lifetimes easily. Maybe when our compilers get as capable as Rust, we can do better in this regard — WhiZTiM, Oct 28 '17 at 11:19

score 1 · Answer 1 · answered Oct 28 '17 at 13:48

No. One of the long list of expressions that can't appear in a constant expression is a new-expression.

The only way to implement a variant and have it be constexpr-friendly is through the use of a union. Although, even with a union, you still wouldn't be able to have a constexpr-friendly variant that can contain a foo since it's not a literal type (by way of it having a non-trivial destructor, by way of std::string having a non-trivial destructor).

C++ constexpr in place aligned storage construction

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