Question

编辑2：程序正在运行。谢谢您的帮助最后一个问题：我在每次推送时都会分配内存。如果我想在pop中删除它，这段代码会起作用吗？

int pop()
{
cout << "About to pop: " << top->getData() << "\n";
  int retVal = top->getData();
  if(top->getLasty()!=NULL)
 {
  node* temp = top;
  top = top->getLasty();
  delete temp;
  cout << "new data is: " << top->getData() << "\n";
 }
return retVal;
}
};

工作代码：

#include <iostream>
using namespace std;

class node
{
  int data;
  node* lasty;

  public:
  node()
  {
   lasty = NULL;
   data = -1;
  }
  node(node* ptr,int dat)
    {
    lasty = ptr;
    data = dat;
    cout << "New node created! Data is " << dat << ", lasty's data is " << ptr->getData() << "\n";
    }
   node* getLasty()
   {
    return lasty;
   }
   int getData()
   {
     return data;
   }
   void setData(int x)
   {
    data = x;
   }

   void setLasty(node* ptr)
   {
    lasty = ptr;
   }
};

class stack
{
node* top;

public:
stack()
{
 cout << "New Stack created!\n";
 top = new node();
}
~stack()
{
delete top;
}

void push(int x)
{
cout << "Before pushing, top's data is " << top->getData() << "\n";
top = new node(top, x);
//node* temp = new node();
//temp->setData(x);
//temp->setLasty(top);
//top = temp;
cout << "TOP'S DATA IS: " << top->getData() << "\n";
cout << "TOP: " << "top lasty's data is: " << top->getLasty()->getData() << "\n\n";
}

int pop()
{
cout << "About to pop: " << top->getData() << "\n";
  int retVal = top->getData();
  if(top->getLasty()!=NULL)
 {
  top = top->getLasty();
  cout << "new data is: " << top->getData() << "\n";
 }
return retVal;
}
};

int main()
{
stack a;
 a.push(1);
 a.push(2);
 a.push(3);
 cout << a.pop();
 cout << a.pop();
    return 0;
}

同样结果我需要更多文字来提交此编辑，因此请考虑这个填充文字。

Answer 1

stack()
{
 cout << "New Stack created!\n";
 node a = node();
 top = &a;
}

节点a是局部变量，因此在传递构造函数体后会被销毁。然后指针top引用被破坏的对象。您应该在堆中创建它并在以后删除它。

stack()
{
 cout << "New Stack created!\n";
 top = new Node();
}

Answer 2

另一个问题：

node(node ptr,int dat)
{

    lasty = &ptr;
    data = dat;
    cout << "New node created! Data is " << dat << ", lasty's data is " << ptr.getData() << "\n";
}

您正在获取复制变量的地址。传递ptr时，编译器正在制作原始项的副本。这称为按值传递。

由于变量的名称是ptr，我假设您正在传递指向节点的指针。您应该指定指针的值，而不是其地址：

// Original
// lasty = &ptr;

// Assigning the pointer value:
lasty = ptr;
// Notice removal of `&`.

您可以指定指针指针。

C ++中Stack实现的奇怪行为

2 个答案: