C ++ constexpr就地对齐存储结构 [英] C++ constexpr in place aligned storage construction

问题描述

我试图创建一个使用std :: aligned_storage来保存数据的对齐变体类型.有没有办法以constexpr的方式构造对象?我读到你不能做新的constexpr放置.

  #include< iostream>#include< string>结构foo{foo(std :: string a，float b):bar1(a)，bar2(b){}std :: string bar1;浮条2;};结构aligned_foo{template< typename ... Args>aligned_foo(Args& ... args){//如何用constexpr构造foo?data_ptr = :: new((void *):: std :: addressof(storage))foo(std :: forward< Args>(args)...);}std :: aligned_storage< sizeof(foo)>贮存;foo * data_ptr;};int main(){aligned_foo("Hello"，0.5);} 

解决方案

否.表达式的一长串列表之一，它们不能出现在常量中表达式是 新表达式 ./p>

实现变体并使它对 constexpr 友好的唯一方法是使用联合.尽管即使有一个联合，您仍然不能拥有一个 constexpr 友好的变体，它可以包含一个 foo ，因为它不是文字类型(通过它具有非平凡的析构函数，例如通过 std :: string 具有非平凡的析构函数).

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);
}

解决方案

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).

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