C ++动态内存//销毁指针指向的内存 [英] C++ Dynamic Memory // Destroying memory pointed by pointer

问题描述

#include <iostream>

using namespace std;

#define SELECT 0

class Z
{
    private:
        int *z1; int *z2;
    public:
        Z(const int x1 = 0, const int x2 = 0);
        Z(const Z &X);
        int *first (void) const {return z1;};
        int *second (void) const {return z2;};
        ~Z(void);
};

Z::Z(const int x1,const int x2){
    z1 = new int(x1);
    z2 = new int(x2);
}

#if SELECT == 1
Z::Z(const Z &X){
    z1 = new int(*X.first() );
    z2 = new int(*X.second() );
}
#else
Z::Z(const Z &X){
    z1 = X.first();
    z2 = X.second();
}
#endif

Z::~Z(){
    delete z1;
    delete z2;
}

int main()
{
 Z *zp;
    zp = new Z(3,5);
    Z  c(0,0);
    c = *zp;
    cout << "Content of c: " << *c.first() << " and " << *c.second() << endl;
    delete zp;
    cout << "Content of c: " << *c.first() << " and " << *c.second() << endl;
}

你好，当我运行这段代码时，我得到类似

Hello, when I run this code, I get something like

Content of c: 3 and 5
Content of c: Garbage and Garbage

我一直期望如此，因为我没有为c创建另一个内存，而是指向zp的内容.但是，当我切换#define SELECT 1时，现在我正在为c创建新的内存.因此，当我删除zp时，c仍应指向正确的值(存储在与zp不同的内存中)，但是我得到的仍然是下面显示的垃圾内容.

I was expecting this since, I am not creating another memory for c instead it points contents of zp. However, when I switch #define SELECT 1, now I am creating new memory for c. So when I delete zp, c still should point correct values (stored in different memory than zp) but what I get is still garbage shown below.

Content of c: 3 and 5
Content of c: Garbage and Garbage

问题出在哪里?

我还有一个问题.当我在VScode中调试此代码时，会出现发生异常".线路的信号未知

I have another question. When I debug this in VScode, I get "Exception Occured. Unknown signal "for the lines

    delete z1;
    delete z2;

在CodeBlocks中，没有错误.有什么问题吗?

In CodeBlocks, there is no error. What is the problem?

感谢您的帮助.

推荐答案

您没有遵循三个 ² .

You don't follow the rule of three ^{1 2}.

复制构造函数只是这三个规则的一部分，但是您还必须实现复制赋值运算符.

The copy constructor is only one part of the rule of three, but you also have to implement the copy assignment operator.

c = * zp; 不会调用副本构造函数，而是调用副本赋值运算符，在您的情况下，这是默认值，因为您没有指定任何内容.

The c = *zp; does not call the copy constructor but the copy assignment operator, which is in your case the default one because you didn't specify any.

默认的复制分配运算符将复制指针，因此在 c = * zp; 之后， c 和 zp 都具有相同的指针 z1 和 z2 会导致对 zp 和对于由 c 分配的内存，内存泄漏.

The default copy assignment operator will copy the pointers, so after c = *zp; both c and zp have the same pointers for z1 and z2 which will lead to a double free (that's most certainly the reason for your last question) of the memory initially allocated by zp and memory leaking for the memory allocated by c.

Z::Z(const Z &X){
    z1 = new int(*X.first() );
    z2 = new int(*X.second() );
}

Z& operator=(const Z& X)
{
  *z1 = *(X.z1);
  *z2 = *(X.z2);

  return *this;
}

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