在堆栈中查找一个项目

Question

在我的代码中，我试图找到放入堆栈的特定项目。为此，我将所有项目移至临时堆栈，将其从原始堆栈中弹出。弹出后，我要将所有项目按原始顺序移回到原始堆栈中。我的代码永远无法识别该项目在堆栈中，因此当它实际上在堆栈中时我发现它没有找到。你能帮我调试我的循环吗？

int main：

#include <iostream>
#include "Stack.h"
#include "Gumball.h"

using namespace std;

int main()
{
  Stack s, gumballStack;
  Gumball g, temp;
  char choice;
  bool choice_flag = true;

  do {
    cin >> choice;
    cin >> g.color;
    switch(choice)
    {
        case 'b':
        case 'B':
            cout << "A" << " " << g.color << " gumball has been bought." << endl << endl;
            g.counter = 0;
            s.isempty();
            s.push(g);
            if(!s.isfull())
                cout << "The gumball is" << " " << g.color << " and has been stored." << endl << endl;
            else
                cout << "There is no room for another gumball." << endl << endl;
            break;
        case 'e':
        case 'E':
            s.isempty();
            temp = s.pop();
            if(s.isempty() && temp.color == g.color)
            {
                cout << "The " << g.color << " gumball has been eaten." << endl << endl;
            }

从这里开始，我相信是错误：

            while(!s.isempty() && g.color != temp.color)
            {
                gumballStack.push(temp);
                g.counter++;
                s.pop();
                cout << " " << temp.counter << endl << endl;
            }
            if(!s.isempty())
            {
                cout << "The " << " " << g.color << " gumball has been eaten." << endl << endl;
            }
            else
            {
                cout << "The gumball cannot be found." << endl << endl;
            }
            while(!gumballStack.isempty())
            {
                //gumballStack.pop();
                s.push(gumballStack.pop());
                gumballStack.pop();
            }
            break;
        case 'q':
        case 'Q':
            choice_flag = false;
            break;
    }
} while(choice_flag);

return 0;
}

.h 文件：

#ifndef STACK_H
#define STACK_H
#include "Gumball.h"

// Interface file  - Stack class definition
class Stack {
    public:
        Stack();
        void push(Gumball);
        Gumball pop();
        bool isempty();
        bool isfull();
    private:
        Gumball gumballs[6+1];
        int top;
};


#endif // STACK_H

回答你的问题 @TOM：

好吧，我认为 .cpp （用于 stack.h）是它将回答您提出的大多数问题：

#include "Stack.h"
#include "Gumball.h"

using namespace std;

// Constructor to initialize the stack
Stack::Stack()
{
   top = -1;
}

// Function to add item x to stack
void Stack::push(Gumball x)
{
   if(!isfull()){
    top++;
    gumballs[top] = x;
    return; }
   else
    return;
}

// Function to remove and return top item of stack
Gumball Stack::pop()
{
    Gumball x;

    if(!isempty()) {
      x = gumballs[top];
      top--;
      return x; }
   else
      return x;
}

// Function to check if stack is empty
bool Stack::isempty()
{
    if (top == -1)
      return true;
    else
      return false;
}

// Function to check if stack is full
bool Stack::isfull()
{
    if (top == 6)
      return true;
    else
      return false;
}

我看到您所说的问题，我多次将 temp 放回到堆栈中......绝对不是我的意图，谢谢您指出这一点。我如何才能将每个口香糖球添加到堆栈中，该口香糖球不等于我正在寻找的物品，而不是相同的物品？

我认为添加 Gumball.h 和 .cpp 将回答您的其他问题，所以这里是：

gumball.h 文件：

#ifndef GUMBALL_H
#define GUMBALL_H
#include <iostream>

using namespace std;

// Interface file  - Gumball class definition
class Gumball
{
    public:
        Gumball();
        string color;
        int counter;
    private: 
};

#endif // GUMBALL_H

gumball.cpp 文件：

#include "Gumball.h"

Gumball::Gumball()
{
    color = " ";
    counter = 0;
}

Answer 1

应该是，

while(!s.isempty() && g.color != temp.color)
        {
            gumballStack.push(temp);
            g.counter++;
            temp = s.pop();  //temp has been updated
            cout << " " << temp.counter << endl << endl;
        }

但请注意，当吃掉的口香糖是堆栈中的最后一个时，此代码将不起作用，因为 s 将为空。

除了同意汤姆的观点并指出您考虑其他解决方案（例如 stl::list ）之外，您应该使用类似的东西

if (s.isempty()) {
    cout << "The gumball cannot be found." << endl << endl;
}
while(!s.isempty()) {
    Gumball temp = s.pop();
    if(temp.color == g.color) {
        cout << "The " << " " << g.color << " gumball has been eaten." << endl << endl;
    } else {
        gumballStack.push(temp);
        g.counter++;
        if (s.isempty()) {
             cout << "The gumball cannot be found." << endl << endl;
        }
    }
}
while(!gumballStack.isempty()) {
      s.push(gumballStack.pop());
      gumballStack.pop();
}

Answer 2

如果没有看到 Stack 的实现，很难说问题出在哪里。但是，由于我发现您的代码的几个部分令人困惑，因此我认为指出其中的位置可能对您有用。如果您更改代码的接口以使其更清晰，那么您的问题可能会变得明显。

// Interface file  - Stack class definition
class Stack {
    public:
        Stack();
        void push(Gumball); //Does this push to the front or back?  
                            //  The stl uses push_back, and push_front, 
                            //  its good to keep this convention 
        Gumball pop();   //Does this pop the front or back?
        bool isempty();  //This function doesn't change Stack, right? 
                         // if so it should be marked const.
        bool isfull();   //Mark as const?
    private:
        Gumball gumballs[6+1];
        int top;
};

在上述问题中，isempty() 的常量性在接下来的内容中尤为重要。

case 'E':
  s.isempty();  //This should be redundent? 
                // isempty is a question, it shouldnt change s.
  temp = s.pop();
  if(s.isempty() && temp.color == g.color)
  {
     cout << "The " << g.color << " gumball has been eaten." << endl << endl;
  }  
  //Here isempty is being used as a question (as if it doesn't change s 
  // - I presume this is the intended use.
  //Also, .color is not a function, it should be, as will be seen. 
  //Also, temp never gets updated in your loop.
  while(!s.isempty() && g.color != temp.color)
  {
    gumballStack.push(temp);  //Why are you pushing multiple copies 
                              // of the same temp
    g.counter++;   //The counter should be an implementation detail, 
        // it should not be exposed like this. 
        // perhaps overload operator++()?
        //Presumably you want g.color to update when you increase the counter? 
        //this doesn't currently happen because g.color is not a function -
        // it points to data.  
        //I'm guessing that When you call color(), 
        // the function should check the value of the counter
        // and obtain the appropriate color.
    s.pop();  //Did you want to update temp here?
    cout << " " << temp.counter << endl << endl;
  }

理想情况下，您将使用迭代器重写整个过程。看一下 std::find 的界面。