Question

我在使用新的连续内存时遇到了一些问题。如果还有其他方法可以指导我。请指导我。
请参考我的代码。

#include <new>  
//================================================
class MyClass
{
 private:
    int ma;
 public:
    MyClass():ma(-1){}      
};
//===========================================

int main()
{
    // I am allocating the memory for holding 10 elements of MyClass on heap
    void* pMyClass = ::operator new(sizeof(MyClass)*10);

    //! Note :: the address of pMyClass1 and pMyClass will now point to same   
    //location after calling placement new  

    MyClass* pMyClass1 = :: new(pMyClass)MyClass();  

    //! Problem with this is that, 
    //! i can only instantiate the constructor for the  base address. That is 
    //!  pMyClass[0]. 
    //! If i have to instantiate it for all the other instances, 
    //! that is pMyClass[1] to pMyClass[9], then how to do it ?
    return 0;
}

Answer 1

你在pMyClass中有内存的开头，而stride是sizeof（MyClass）。所以，你需要做的是例如：

MyClass* pMyClass2 = ::new((MyClass*)pMyClass + 1)MyClass();

Answer 2

你必须在循环中调用placement new，迭代10个连续的内存块：

int main()
{
    void* pMyClass = ::operator new(sizeof(MyClass)*10);
    MyClass* pMyClass1 = reinterpret_cast<MyClass*>(pMyClass);
    for (size_t i=0; i<10; ++i) {
        ::new(pMyClass1++)MyClass();
    }
    // ... 
}

Answer 3

尝试：

MyClass* pMyClass1 = :: new(pMyClass)MyClass[10];

为连续内存放置新内容

3 个答案: