什么时候使用拷贝构造函数？

Question

我创建了一个 class，它有一个带参数的构造函数。我认为我在这里需要一个复制构造函数是对的吗？

我有一个class，不接受指点：

class xyz : public Node
{
public:
    xyz( uint8_t node);
    xyz(const xyz& cxyz);
    ~xyz();

private:
    uint8_t m_node;
    uint16_t m_gainxyz;

和基地：

class Node
{
public:

Node();
virtual ~Node();

protected:
    std::string  m_name;

当我这样做时：

xyz xyz = Initxyz(node);

编译器告诉我创建一个拷贝构造函数。

其中：

xyz PD::Initxyz(Source& inputNode)
{
    if (inputNode.getNodeNumber() > 10 )
        {
        xyz element(inputNode.getNodeNumber());
        element.setInputNode(inputNode);
        return element; 
        }
    else
        {
        std::cout << "ERROR IN XYZ CONFIGURATION" << std::endl;
        //Throw Exception;
        }
}

但是根据我在网上看到的：

If the object has no pointers to dynamically allocated memory, a shallow copy is probably sufficient. Therefore the default copy constructor, default assignment operator, and default destructor are ok and you don't need to write your own.

http://www.fredosaurus.com/notes-cpp/oop-condestructors/copyconstructors.html

如果我使用带参数的构造函数，我是否也必须有一个？

Answer 1

我会提供一些规范的参考，以便正式地解释这件事。

如果我正确理解你想要什么，当你需要复制你的对象时 N4296::12.8/1 [class.copy]:

A class object can be copied or moved in two ways: by initialization (12.1, 8.5), including for function argument passing (5.2.2) and for function value return (6.6.3); and by assignment (5.18).

实际上，复制构造函数可能被隐式声明为已删除，因此如果您在需要复制时尝试调用它，将会出现编译时错误。

相关参考：

An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/move constructor for a class X is defined as deleted (8.4.3) if X has:

(11.1) — a variant member with a non-trivial corresponding constructor and X is a union-like class,

(11.2) — a potentially constructed subobject type M (or array thereof) that cannot be copied/moved because overload resolution (13.3), as applied to M’s corresponding constructor, results in an ambiguity or a function that is deleted or inaccessible from the defaulted constructor,

(11.3) — any potentially constructed subobject of a type with a destructor that is deleted or inaccessible from the defaulted constructor, or,

(11.4) — for the copy constructor, a non-static data member of rvalue reference type.

其中潜在构造子对象的定义如下N4296::12/5

For a class, its non-static data members, its non-virtual direct base classes, and, if the class is not abstract (10.4), its virtual base classes are called its potentially constructed subobjects.

Answer 2

I have created a class, which has a constructor that takes a parameter. Am I right in thinking that I need a copy constructor here?

您不需要为您的类提供复制构造函数。编译器生成的就足够了。

这里是a working example based on your code:

#include <iostream>
#include <string>
#include <stdint.h>

class Node
{
public:

  Node() {}
  virtual ~Node() {}

protected:
  std::string  m_name;
};

class xyz : public Node
{
public:
    xyz( uint8_t node) {}

private:
    uint8_t m_node;
    uint16_t m_gainxyz;
};

xyz Initxyz(Node)
{
    return xyz(42);
}

int main()
{
    Node node;
    xyz xyz = Initxyz(node);
}

Answer 3

当你想做深拷贝而不是浅拷贝时需要复制构造函数copy.It意味着你正在避免悬空指针问题。实际上编译器也提供了默认的复制构造函数，但如果你没有任何字符串到构造函数进行复制，它就可以正常工作。但是如果你想将字符串从 class 的一个对象复制到 class 的另一个对象，那么你不能将一个内存地址指向两个指针。尝试做拷贝构造函数的例子

什么时候使用拷贝构造函数？

When to use a copy constructor?

c++

constructor