Question

我想在不调用构造函数的情况下在堆栈上分配类数组。以下示例澄清：

template<class t,int SetNum> class set
{
private:
    t Storage[SetNum];
};

class myClass
{
private:
    int* Array;
public:
    myClass()
    {
        Array=new int[10];
    }
}
int main()
{
    set<myClass,10> Set;
}

我不想为调用Array的构造函数时发生的myClass分配10个新的整数，但仍希望为myClass分配空间。

Answer 1

您必须有一个unsigned char（或类似）数组用作元素的“后备存储”，然后调用展示位置new运算符（请参阅例如here ）在那里构建你的实例（顺便说一句，这是std::vector已经做过的事情。）

警告： 如果您使用placement new，您有责任手动释放您使用它创建的对象，显式调用析构函数;另外，传递给placement new的指针必须与你正在创建的对象正确对齐，否则可能会发生不好的事情。

另见this question。

使用所描述的技术构建的std::array和std::vector的扭曲混合示例（要求C ++ 11使union技巧起作用）：

#include <cstddef>
#include <memory>
#include <stdexcept>
#include <iostream>

template<typename T, std::size_t N>
class array_noinit
{
    union storage_helper
    {
    private:
        // this member assures that the storage is properly aligned
        T t;
    public:
        unsigned char storage[sizeof(T)*N];

        // needed because T's constructor/destructor is implicitly deleted
        storage_helper() { };
        ~storage_helper() { };
    };

    storage_helper s;

    std::size_t _size;
    T * _storage;
public:
    array_noinit() :
        _size(0), _storage((T*)s.storage)
    {}

    ~array_noinit()
    {
        while(_size>0)
            pop_back();
    }

    void push_back(const T & elem)
    {
        if(_size>=N)
            throw std::runtime_error("Not enough capacity.");
        new(_storage+_size) T(elem);
        _size++;
    }

    void pop_back()
    {
        if(_size>0)
        {
            _size--;
            _storage[_size].~T();
        }
    }

    T & at(std::size_t Idx)
    {
        if(Idx>=_size)
            throw std::out_of_range("Idx out of range.");
        return _storage[Idx];
    }

    std::size_t size() const
    {
        return _size;
    }

    std::size_t capacity() const
    {
        return N;
    }
};

class A
{
    int _i;
public:
    A(int i) : _i(i)
    {
        std::cout<<"In A constructor - "<<_i<<"\n";
    }

    A(const A & right)
        : _i(right._i)
    {
        std::cout<<"In A copy constructor - "<<_i<<"\n";
    }

    ~A()
    {
        std::cout<<"In A destructor - "<<_i<<"\n";
    }
};

int main()
{
    std::cout<<"With ints\n";
    array_noinit<int, 4> arr;
    arr.push_back(1);
    arr.push_back(2);
    arr.push_back(3);
    arr.push_back(4);

    for(std::size_t i=0; i<4; i++)
        std::cout<<arr.at(i)<<" ";
    std::cout<<"\n";

    std::cout<<"With a class - we don't add anything\n";
    array_noinit<A, 10> arr2;

    std::cout<<"With a class - now we add stuff\n";
    array_noinit<A, 10> arr3;
    arr3.push_back(A(1));
    arr3.push_back(A(2));
    arr3.push_back(A(3));
    return 0;
}

输出：

With ints
1 2 3 4 
With a class - we don't add anything
With a class - now we add stuff
In A constructor - 1
In A copy constructor - 1
In A destructor - 1
In A constructor - 2
In A copy constructor - 2
In A destructor - 2
In A constructor - 3
In A copy constructor - 3
In A destructor - 3
In A destructor - 3
In A destructor - 2
In A destructor - 1

编辑有一个much simpler way to get aligned storage。

未初始化类的数组

1 个答案: