Question

这个程序应该使用一个后缀算术表达式，然后编译该表达式的值。每次读取一个整数，它将被推入堆栈。否则，如果+， - ，*是，则会弹出两个整数读取。

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;    
}

当我在main中调用print函数时，我得到一个无限循环作为输出。我试着寻找导致无限循环的东西，但我找不到它。

Answer 1

我看到的问题：

在delete中使用push()：

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

未在delete中使用pop()：

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;
}

你需要：

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
   }
}

在delete中使用print()。

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

缺少用户定义的析构函数。您需要删除对象中的对象。否则，你正在泄漏记忆。以下代码中的某些内容应该有效。
```
Stack::~Stack()
{
   while (head)
   {
      Node * temp = head;
      head = head->next;
      delete temp;
   }
}
```

Answer 2

以下是push和pop的建议。

尝试理解逻辑。

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;
}

此外，对于代码中的每个数字，所有if / else都可以执行以下操作：

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

调用堆栈打印功能时无限循环

2 个答案: