我在检查aligned_storage中cppref，但我认为它的例子是越野车。这是代码：

#include <iostream>
#include <type_traits>
#include <string>
 
template<class T, std::size_t N>
class static_vector
{
    // properly aligned uninitialized storage for N T's
    typename std::aligned_storage<sizeof(T), alignof(T)>::type data[N];
    std::size_t m_size = 0;
 
public:
    // Create an object in aligned storage
    template<typename ...Args> void emplace_back(Args&&... args) 
    {
        if( m_size >= N ) // possible error handling
            throw std::bad_alloc{};
 
        // construct value in memory of aligned storage
        // using inplace operator new
        new(&data[m_size]) T(std::forward<Args>(args)...);
        ++m_size;
    }
 
    // Access an object in aligned storage
    const T& operator[](std::size_t pos) const 
    {
        // note: needs std::launder as of C++17
        return *reinterpret_cast<const T*>(&data[pos]);
    }
 
    // Delete objects from aligned storage
    ~static_vector() 
    {
        for(std::size_t pos = 0; pos < m_size; ++pos) {
            // note: needs std::launder as of C++17
            reinterpret_cast<T*>(&data[pos])->~T();
        }
    }
};
 
int main()
{
    static_vector<std::string, 10> v1;
    v1.emplace_back(5, '*');
    v1.emplace_back(10, '*');
    std::cout << v1[0] << 'n' << v1[1] << 'n';
}

在这里，我们要创建一个使用放置 new的静态向量。问题是typename std::aligned_storage<sizeof(T), alignof(T)>::type类型是 POD，而不是T. 所以我们需要在使用前投射它。我认为代码应该是这样的：

#include <iostream>
#include <type_traits>
#include <string>
 
template<class T, std::size_t N>
class static_vector
{
    // properly aligned uninitialized storage for N T's
    typename std::aligned_storage<sizeof(T), alignof(T)>::type data[N];
    T* data_ptr = reinterpret_cast<T*>(data);

    std::size_t m_size = 0;
 
public:
    // Create an object in aligned storage
    template<typename ...Args> void emplace_back(Args&&... args) 
    {
        if( m_size >= N ) // possible error handling
            throw std::bad_alloc{};
 
        // construct value in memory of aligned storage
        // using inplace operator new
        new(&data_ptr[m_size]) T(std::forward<Args>(args)...);
        ++m_size;
    }
 
    // Access an object in aligned storage
    const T& operator[](std::size_t pos) const 
    {
        // note: needs std::launder as of C++17
        return *reinterpret_cast<const T*>(&data_ptr[pos]);
    }
 
    // Delete objects from aligned storage
    ~static_vector() 
    {
        for(std::size_t pos = 0; pos < m_size; ++pos) {
            // note: needs std::launder as of C++17
            reinterpret_cast<T*>(&data_ptr[pos])->~T();
        }
    }
};
 
int main()
{
    static_vector<std::string, 10> v1;
    v1.emplace_back(5, '*');
    v1.emplace_back(10, '*');
    std::cout << v1[0] << 'n' << v1[1] << 'n';
}

我对吗？虽然我不知道为什么原始代码在 clang 中有效。

更新：

让我说得更具体一点。按照标准，aligned_storage类型是 POD 而不是T。所以它的实现可以如下：

template<std::size_t Len, std::size_t Align /* default alignment not implemented */>
struct aligned_storage {
    struct type {
        alignas(Align) unsigned char data[Len];
    };
};

现在，如果您使用 访问它data[pos]，地址将根据unsigned charsize=1增加，不是sizeof(T)吗？

回答

typename std::aligned_storage<sizeof(T), alignof(T)>::type data[N];