C++ 中重载 = 运算符当有要复制的值数组时

发布于 2024-10-04 00:52:22 字数 900 浏览 6 评论 0原文

我对 C++ 有点陌生，所以我想这是一个非常基本的问题。

假设我有这个类：

// file Graph.h
class Graph { 
public:
  Graph(int N); // contructor
  ~Graph();     // destructor  
  Graph& operator=(Graph other);
private:
  int * M;
  int N;
};

// file Graph.cpp
Graph :: Graph(int size) {
  M = new int [size];
  N = size;
}

Graph :: ~Graph() {
  delete [] M;
}

我想创建一个赋值运算符，它将复制数组 M[] 的内容，但当我在复制后更改它时不覆盖它（我认为这是通过不复制实际指针而仅复制内容来完成的，不知道我是否正确）。这是我尝试过的：

Graph& Graph::operator=(Graph other) {
  int i;
  N = other.N;
  M = new int [N];
  for (i = 0; i < N; i++)
    M[i] = other.M[i];
  return *this;
 }

这是正确的吗？还有其他方法可以做到这一点吗？

编辑：我忘记了一个重要的问题。为什么我必须像 Graph& 那样声明它operator=(Graph other); 而不仅仅是： Graph operator=(Graph other); 这是我的书（C++: The Complete Reference，第二版，Herbert Schildt，第 355-357 页）？

原文

I'm somewhat new to C++ so, I guess this is a very basic question.

Suppose I have this class:

// file Graph.h
class Graph { 
public:
  Graph(int N); // contructor
  ~Graph();     // destructor  
  Graph& operator=(Graph other);
private:
  int * M;
  int N;
};

// file Graph.cpp
Graph :: Graph(int size) {
  M = new int [size];
  N = size;
}

Graph :: ~Graph() {
  delete [] M;
}

I want to create an assignment operator that will copy the contents of array M[] but not to overwrite it when I change it after the copy (I think this is accomplished by not copying the actual pointer but only the content, don't know if I'm right). This is what I've tried:

Graph& Graph::operator=(Graph other) {
  int i;
  N = other.N;
  M = new int [N];
  for (i = 0; i < N; i++)
    M[i] = other.M[i];
  return *this;
 }

Is this correct? Are there other ways to do this?

edit: An important question I forgot. Why I must declare it like Graph& operator=(Graph other); and not just: Graph operator=(Graph other); which is what's written in my book (C++: The Complete Reference, 2nd ed, Herbert Schildt, pages 355-357)?

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╰つ倒转 2024-10-11 00:52:22

规范的方法是使用 std::vector 来避免自己管理内存。不过，对于这个练习，做你想做的事情的正确方法是：

#include <algorithm>

class Graph
{
public:    
    Graph(size_t n) { data_ = new int[n](); size_ = n; }

    Graph(Graph const& g)
    {
        data_ = new int(g.size_);
        size_ = g.size_;
        std::copy(g.data_, g.data_ + g.size_, data_);
    }

    ~Graph() { delete[] data_; }

    void swap(Graph& g) throw()
    {
        std::swap(data_, g.data_);
        std::swap(size_, g.size_);
    }

    Graph& operator=(Graph g) { g.swap(*this); return *this; }

private:
    int* data_;
    size_t size_;
};

谷歌“复制和交换习惯用法”来了解代码背后的基本原理。请注意，您的分配运算符会泄漏内存（原始数组被覆盖但从未删除），如果分配失败，您最终会得到一个损坏的对象。此外，x = x 不会执行预期的操作。这三个陷阱很常见，以复制和交换方式编写赋值运算符可以避免它们。

编辑：对于您的其他问题，返回引用允许您链接赋值，例如 a = b = c，这对于内置类型有效。它可能是也可能不是您想要的（通常是）。

The canonical way is to use a std::vector<int> to avoid managing memory yourself. For the exercise though, the right way to do what you want is:

#include <algorithm>

class Graph
{
public:    
    Graph(size_t n) { data_ = new int[n](); size_ = n; }

    Graph(Graph const& g)
    {
        data_ = new int(g.size_);
        size_ = g.size_;
        std::copy(g.data_, g.data_ + g.size_, data_);
    }

    ~Graph() { delete[] data_; }

    void swap(Graph& g) throw()
    {
        std::swap(data_, g.data_);
        std::swap(size_, g.size_);
    }

    Graph& operator=(Graph g) { g.swap(*this); return *this; }

private:
    int* data_;
    size_t size_;
};

Google "copy and swap idiom" for the rationale behind the code. Note that your assigment operator leaks memory (the original array is overwritten but never deleted) and if the allocation fails, you end up with a broken object. Moreover, x = x won't do what it is expected to do. These three pitfalls are common, and writing assignment operators in copy-and-swap style avoids them.

EDIT: For your other question, returning a reference allows you to chain assignments, like a = b = c, which is valid for built in types. It may or may not be what you want (it usually is).

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