将STD：：Memcpy用于非日常可复制类型的对象

Question

#include <iostream>
#include <cstring>

using std::cout;
using std::endl;

struct X
{
    int a = 6;
    X(){ }
    X(const X&)
    {
        cout << "X()" << endl;
    }
};

X a;
X b;
int main()
{
    a.a = 10;
    std::memcpy(&b, &a, sizeof(X));
    cout << b.a << endl; //10
}

Answer 1

你问：

如果我们将该函数应用于一个非可复制类型的对象，我们可能会遇到什么潜在的问题？

下面是一个非常简单的示例，它说明了对非可复制类型的对象使用std::memcpy的问题。

#include <cstring>

struct A
{
   A(int size) : size_(size), data_(new int[size]) {}
   ~A() { delete [] data_; }

   // The copy constructor and the copy assignment operator need
   // to be implemented for the class too. They have been omitted
   // to keep the code here minimal.

   int size_;
   int* data_;
};

int main()
{
   A a1(10);
   A a2(20);
   std::memcpy(&a1, &a2, sizeof(A));

   // When we return from the function, the original data_ of a1
   // is a memory leak. The data_ of a2 is deleted twice.

   return 0;
}