后缀评估

Question

我正在编写一段代码，对给定的后缀表达式求值。每个操作数和运算符之间用空格分隔，最后一个运算符后跟空格和“x”。

例子：

内插表达式：（2*3 4)*(4*3 2)

后缀表达式：23*443*2*x

“x”暗示表达的结束。

输入（后缀表达式）作为字符串由另一个函数给出，该函数将内缀表达式转换为后缀表达式。

后缀评估的功能是：

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;
}

我给出的例子的输出应该是“140”，但我得到的是“6”。请帮我找出错误。

推送和弹出方法如下（以防有人想复习）：

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

尝试与以下代码进行比较。

#include<iostream>
using namespace std;
#include<conio.h>
#include<string.h>
#include<math.h>

class A
{
    char p[30],ch;
    int i,top,s[30],y1,y2,x,y,r;

public:
    A()
    {
        top=-1;
        i=0;
    }
    void input();
    char getsymbol();
    void push(int);
    int pop();
    void evaluation();
};

void A :: input()
{
    cout<<"enter postfix expression\n";
    gets(p);
}
char A :: getsymbol()
{
    return p[i++];
}
void A :: push(int ch)
{
    if(top==29)
        cout<<"stack overflow\n";
    else
        s[++top]=ch;
}
int A :: pop()
{
    if(top==-1)
    {
        cout<<"stack underflow\n";
        return 0;
    }
    else
        return s[top--];
}
void A :: evaluation()
{
    ch=getsymbol();
    while(ch!='\0')
    {
        if(ch>='a'&&ch<='z'||ch>='A'&&ch<='Z')
        {
            cout<<"enter the value for "<<ch;
            cin>>x;
            push(x);
        }
        if(ch=='+'||ch=='-'||ch=='*'||ch=='/'||ch=='^')
        {
            y2=pop();
            y1=pop();
            if(ch=='+')
            {
                y=y1+y2;
                push(y);
            }
            else if(ch=='-')
            {
                y=y1-y2;
                push(y);
            }
            else if(ch=='^')
            {
                y=y1^y2;
                push(y);
            }
            else if(ch=='*')
            {
                y=y1*y2;
                push(y);
            }
            else if(ch=='/')
            {
                y=y1/y2;
                push(y);
            }
            else
            {
                cout<<"entered operator has no value\n";
            }    
        }
        ch=getsymbol();
    }
    if(ch=='\0')
    {
        r=pop();
        cout<<"the result is "<<r;
    }
}
int main()
{
    A a;
    int m=0;
    while(m==0)
    {
        a.input();
        a.evaluation();
        cout<<"enter 0 to continue 1 to exit\n";
        cin>>m;
    }
    getch();
    return 0;
}

Answer 2

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

一旦a）当前字符是空格，或者b）下一个字符是x，您就退出这个循环。当前字符在处理过程中很早就变成了空格；您只处理字符串的一小部分。