c++ - 后缀评估

Question

我正在编写一个代码来评估给定的后缀表达式。每个操作数和运算符由一个空格分隔，最后一个运算符后跟一个空格和一个“x”。

示例:

中缀表达式:(2*3+4)*(4*3+2)

后缀表达式:2 3 * 4 + 4 3 * 2 + * x

“x”表示表达式结束。

输入(后缀表达式)由另一个将中缀表达式转换为后缀表达式的函数作为字符串给出。

后缀求值的函数是:

int pfeval(string input)
{
int answer, operand1, operand2, i=0;
char const* ch = input.c_str();
node *utility, *top;
utility = new node;
utility -> number = 0;
utility -> next = NULL;
top = new node;
top -> number = 0;
top -> next = utility;

while((ch[i] != ' ')&&(ch[i+1] != 'x'))
{
    int operand = 0;
    if(ch[i] == ' ') //to skip a blank space
        i++;
    if((ch[i] >= '0')&&(ch[i] <= '9')) //to gather all digits of a number
    {
        while(ch[i] != ' ')
        {
            operand = operand*10 + (ch[i]-48);
            i++;
        }
        top = push(top, operand);
    }
    else
    {
        top = pop(top, operand1);
        top = pop(top, operand2);
        switch(ch[i])
        {
        case '+': answer = operand2 + operand1;
        break;
        case '-': answer = operand2 - operand1;
        break;
        case '*': answer = operand2 * operand1;
        break;
        case '/': answer = operand2 / operand1;
        break;
        case '^': answer = pow(operand2, operand1);
        break;
        }
        top = push(top, answer);
    }
    i++;
}
pop(top, answer);
cout << "\nAnswer: " << answer << endl;
return 0;
}

我给出的示例的输出应该是“140”，但我得到的是“6”。请帮我找出错误。

push和pop方法如下(万一有人要复习):

class node
{
public:
int number;
node *next;
};

node* push(node *stack, int data)
{
node *utility;
utility = new node;
utility -> number = data;
utility -> next = stack;
return utility;
}

node* pop(node *stack, int &data)
{
node *temp;
if (stack != NULL)
{
    temp = stack;
    data = stack -> number;
    stack = stack -> next;
    delete temp;
}
else cout << "\nERROR: Empty stack.\n";
return stack;
}

Answer 1

while((ch[i] != ' ')&&(ch[i+1] != 'x'))

一旦 a) 当前字符是空格，或 b) 下一个字符是“x”，您就退出此循环。当前角色在此过程的早期就变成了一个空格；您只处理字符串的一小部分。