調用堆棧打印函數時無限循環

Question

該程序應該採用後綴算術表達式，然後編譯該表達式的值..每次讀取整數時，它將被推入堆棧中。否則，兩個整數將是如果讀取了+， - ，*，則彈出。

class Stack { 
    Node *head; 

public: 
    Stack() { 
     head = NULL; 
    }; 

    void push(int data); 
    int pop(); 
    bool isEmpty(); 
    void print(); 
}; 

void Stack::push(int data) 
{ 
    Node * temp = new Node(data); 
    temp->next = head; 
    head = temp; 
    delete temp; 
} 

int Stack::pop() 
{ 
    int x = head->data; 
    head = head->next; 
    return x; 
} 

bool Stack::isEmpty(){ 
    return head == NULL; 
} 

void Stack::print(){ 
    Node * temp = head; 
    while (temp != NULL){ 
     cout << temp->data << " "; 
     temp = temp->next; 
    } 
    delete temp; 
} 


int main() { 

    Stack st; 
    char exp [] = "23+", c; 
    int i, a; 

    for (i = 0; exp[i] != '\0'; i++){ 
     c = exp[i]; 

     if (c == '+'&&!st.isEmpty()){ 
      a = st.pop() + st.pop(); 
      st.push(a); 
     } 
     else if (c == '-'&&!st.isEmpty()){ 
      a = st.pop() - st.pop(); 
      st.push(a); 
     } 
     else if (c == '/'&&!st.isEmpty()){ 
      a = st.pop()/st.pop(); 
      st.push(a); 
     } 
     else if (c == '*'&&!st.isEmpty()){ 
      a = st.pop() * st.pop(); 
      st.push(a); 
     } 
     else if (c == '0') 
      st.push(0); 
     else if (c == '1') 
      st.push(1); 
     else if (c == '2') 
      st.push(2); 
     else if (c == '3') 
      st.push(3); 
     else if (c == '4') 
      st.push(4); 
     else if (c == '5') 
      st.push(5); 
     else if (c == '6') 
      st.push(6); 
     else if (c == '7') 
      st.push(7); 
     else if (c == '8') 
      st.push(8); 
     else if (c == '9') 
      st.push(9); 

     cout << c << endl; 
     st.print(); 
    } 
    cin >> a; 
    return 0;  
} 

當我打電話主打印功能，我得到一個無限循環作爲輸出.. 我試圖尋找這是造成一個無限循環的事情，但我找不到它。

Answer 1

問題，我看到：

1. 使用deletepush()：

   void Stack::push(int data) 
   { 
       Node * temp = new Node(data); 
       temp->next = head; 
       head = temp; 
       delete temp; // This needs to go. 
   } 
   

2. 在pop()不使用delete：

   int Stack::pop() 
   { 
       // Problem 1. 
       // What if head is NULL? 
   
       int x = head->data; 
   
       // Problem 2 
       // The old value of head is gone. It's a memory leak. 
       head = head->next; 
       return x; 
   } 
   

   您需要：

   int Stack::pop() 
   { 
       if (head != NULL) 
       { 
        int x = head->data; 
        Node * temp = head; 
        head = head->next; 
        delete temp; 
        return x; 
       } 
       else 
       { 
        // Figure out what you want to do if head is NULL 
       } 
   } 
   

3. 在print()使用delete。

   void Stack::print(){ 
       Node * temp = head; 
       while (temp != NULL){ 
        cout << temp->data << " "; 
        temp = temp->next; 
       } 
       delete temp; // This needs to go. 
   } 
   

4. 缺少用戶定義的析構函數。您需要刪除對象中的對象。否則，你正在泄漏記憶。下面的代碼行應該工作。

   Stack::~Stack() 
   { 
       while (head) 
       { 
        Node * temp = head; 
        head = head->next; 
        delete temp; 
       } 
   } 
   

Answer 2

這是對push和pop的建議。

試圖理解邏輯。

void Stack::push(int data) 
{ 
    Node * temp = new Node(data); 
    temp->next=head; 
    head=temp; 
    //Do not delete temp; deleting temp will delete the new added Node 
} 

int Stack::pop() 
{ 
    Node* temp = Head; 
    int x=head->data; 
    head=head->next; 
    delete temp; //here you free the released memory. 
    return x; 
} 

而且，而是所有的if/else，代碼中的每一個數字，你可以做如下：

else if(c>=0 && c<=9){ 
    st.push(c-'0'); 
}