使用堆栈C的中缀到后缀：错误代码6

Question

我的任务是使用单链表实现堆栈，将中缀形式的字符串转换为后缀形式。为简单起见，此字符串不包含任何空格。

简而言之，我的算法是：

>

按操作顺序使用字符及其关联优先级创建临时节点

如果是操作而不是数字，则将其推到堆栈上/如果是数字，则自动将其附加到后缀字符串

每次将一个字符推送到堆栈上时，如果堆栈的顶部节点的优先级高于下一个字符的临时节点，请将其从堆栈中弹出，并将其附加到后缀字符串中。

这些步骤在手动执行中缀到后缀时有效。每当我试图运行我的代码，我总是得到一个错误6 SIGABRT。我的代码应该很容易理解。谁能告诉我这个错误是什么意思，为什么我会得到它，以及如何修复它，以便我的代码正确地输出后缀字符串？

#include <iostream>
#include <string>

using namespace std;
string postfix; //infix and postfix strings

//function to return true if a character is an operation
bool isoperator(char a)
{
    if(a == '(' ||  ')' || '*' || '/' || '+' || '-') //might need to 
change to "" instead of ''
    {
        return(true);
    }
    else
    {
        return(false);
    }
}

//node class
class node
{
public:
    char character;
    //double number;
    int level; //to check for precedence of operations
    node *ptr;
    void assignlevel()
    {
        switch(character)
        {
            case ')':
                level = 3;
                break;
            case '(':
                level = 0;
                break;
            case '+':
                level = 1;
                break;
            case '-':
                level = 1;
                break;
            case '*':
                level = 2;
                break;
            case '/':
                level = 2;
                break;
            default:
                level = 0;
        }
}
friend class stack;
};

//stack class
class stack
{
public:
    node *top, *temp;

//Constructor Function
stack()
{
    top = new node;
    top->character = '&';
    top->ptr = NULL;
    top->level = 0;
    temp = new node;
    temp->ptr = NULL;
}

//Empty
bool empty()
{
    return(top->character == '&');
}

//Read character from string
void readchar(char a)
{
    temp->character = a;
    temp->assignlevel();
}

//Check Precedence of top and temp
bool precedence()
{
    return(top->level >= temp->level);
}

//Push function for infix to postfix
void push1(char a)
{
    readchar(a);
    if(isoperator(temp->character)) //Push onto stack if character is an operation
    {
        if(empty())
        {
            top->character = temp->character;
            top->assignlevel();
        }
        else
        {
            node *v = new node;
            v->character = temp->character;
            v->level = temp->level;
            v->ptr = top;
            top = v;
            delete v;
        }
    }
    else //append to string if character is number
    {
        postfix += temp->character;
    }

    if(precedence()) //we check if we have to pop every time we push 
onto the stack
        {
            pop1();
        }
    }

    void pop1() //Pop onto postfix string
    {
        postfix += top->character;
        node *w = top->ptr;
        delete &top;
        top = w;
        delete w;
    }
};

int main()
{
    string infix = "2+3-5*(7+1)";
    stack op;
    for(int i = 0; i < infix.size(); ++i)
    {
        op.push1(infix[i]);
    }


    for(int j = 0; j < infix.size(); j++)
    {
        cout << postfix[j];
    }
    return 0;
}

Answer 1

为什么在推送中要“删除v”？这将删除刚刚创建的节点。