将 std::memcpy 用于非平凡可复制类型的对象 [英] Using std::memcpy to object of non-trivially copyable type

问题描述

标准定义我们可以通过以下方式使用 std::memcpy int :

The standard defines we can use std::memcpy int the following way:

对于任何可简单复制的类型 T，如果两个指向 T 的指针指向不同的 T 对象 obj1 和 obj2，其中 obj1 和 obj2 都不是基类子对象，如果构成 obj1 的底层字节 (1.7) 是复制到 obj2 中，obj2 随后将保持与 obj1 相同的值.

For any trivially copyable type T, if two pointers to T point to distinct T objects obj1 and obj2, where neither obj1 nor obj2 is a base-class subobject, if the underlying bytes (1.7) making up obj1 are copied into obj2, obj2 shall subsequently hold the same value as obj1.

如果我们将该函数应用于非平凡可复制类型的对象，我们可能会遇到什么潜在问题?下面的代码就好像它适用于可简单复制的类型一样:

What potential problem we could get if we applied that function to an object of non-trivially copyable type? The following code works as if it worked for trivially-copyable type:

#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
}

演示

推荐答案

你问:

如果我们将该函数应用于非平凡可复制类型的对象，我们可能会遇到什么潜在问题?

What potential problem we could get if we applied that function to an object of non-trivially copyable type?

这是一个非常简单的例子，它说明了将 std::memcpy 用于非平凡可复制类型的对象的问题.

Here's a very simple example that illustrates the problem of using std::memcpy for objects of non-trivially copyable type.

#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;
}

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