后缀评估

Postfix evaluation

本文关键字：评估后缀更新时间：2023-10-16

我正在编写一个计算给定后缀表达式的代码。每个操作数和运算符由空格分隔，最后一个运算符后跟一个空格和一个"x"。

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"表示表达式的结束。

输入（后缀表达式）由另一个函数以字符串形式给出，该函数将中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;
}

我给出的示例的输出应该是"140"，但我得到的是"6"。请帮我找到错误。

推送和弹出方法如下（以防有人想要审查）：

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;
}

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"，您就可以退出此循环。当前角色在这个过程的早期就变成了一个空间;只处理字符串的一小部分。

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 expressionn";
    gets(p);
}
char A :: getsymbol()
{
    return p[i++];
}
void A :: push(int ch)
{
    if(top==29)
        cout<<"stack overflown";
    else
        s[++top]=ch;
}
int A :: pop()
{
    if(top==-1)
    {
        cout<<"stack underflown";
        return 0;
    }
    else
        return s[top--];
}
void A :: evaluation()
{
    ch=getsymbol();
    while(ch!='')
    {
        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 valuen";
            }    
        }
        ch=getsymbol();
    }
    if(ch=='')
    {
        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 exitn";
        cin>>m;
    }
    getch();
    return 0;
}