补缀到后缀计算器/阅读输入

Infix to postfix calculator/ reading input

本文关键字：输入 计算器 后缀      更新时间：2023-10-16

我需要使用C++中的堆栈和队列来实现中缀到后缀计算器。我知道该遵循什么算法，但我在启动时遇到了问题。我的程序应该如下所示：

Enter a valid infix expression: 3 + 4 * 5 / 6
The resulting postfi expression is: 3 4 5 * / 6 +
The result is: 6.3333

所以我需要从命令行读取输入，这就是我的问题所在。这是我迄今为止的代码：

using namespace std;
#include <iostream>
#include <stdlib.h>
#include <stack>
#include <queue>
int main() {
  stack <string> convert;
  stack <string> evaluate;
  queue <string> store;
  string data;
  float num;
  cout << "Enter a valid infix expression: ";
  while (getline(cin, data)) {
    store.push(data);
  }
return 0;
}

我的问题是如何停止循环，以及如何从字符串输入中获取数字，以便将它们推入队列以便稍后打印？我的代码将整个字符串推入队列中的第一个插槽。希望有人能帮忙。

感谢

使用我编写的类工具

Tools.h

static class Tools
{
   public:
       static char* toPostfix(char* source);
       static double calculatePostfix(char* source);
       static bool contain(char* source,char character);
 };

Tools.cpp

#include "Tools.h"
#include <stack>
#include <iostream>
#include <string>
using namespace std;
char* Tools::toPostfix(char* source)
{
    stack<char> symbol;
    string postfix = "";
    int i = 0;
    char variables[] = { "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ" };
    bool find = false;
    while (*source != '')
    {
        switch (*source)
        {
        case '+':
        case '-':
        case '*':
        case '/':
            symbol.push(*source); 
            break;
        case ')':
            postfix += symbol.top();
            symbol.pop();
        default:
            find = Tools::contain(variables, *source);
            if (find)
            {
                 postfix += *source;
                 find = false;
             }
        }
        source++;
    }
    // attach other operator in stack(Pop All)
    while (!symbol.empty())
    {
         postfix += symbol.top();
         symbol.pop();
     }
     char* p = new char(postfix.length());
     const char* o = postfix.c_str();
     for (int i = 0; i < postfix.length(); i++)
         p[i] = o[i];
     return p;
}
double Tools::calculatePostfix(char* source)
{
    char numbers[] = { "0123456789" };
    stack<double> number;
    for (int i = 0; i < strlen(source); i++)
    {
        char character = source[i];
        if (Tools::contain(numbers, character))
        {
            number.push(atof(&character));
        }
        else
        {
            double number1, number2;
            switch (character)
            {
            case '+':
                number2 = number.top();
                number.pop();
                number1 = number.top();
                number.pop();
                number.push(number1 + number2);
                break;
            case '-':
                number2 = number.top();
                number.pop();
                number1 = number.top();
                number.pop();
                number.push(number1 - number2);
                break;
            case '*':
                number2 = number.top();
                number.pop();
                number1 = number.top();
                number.pop();
                number.push(number1 * number2);
                break;
            case '/':
                number2 = number.top();
                number.pop();
                number1 = number.top();
                number.pop();
                number.push(number1 / number2);
                break;
            }
        }
    }
    return number.top();
}
bool Tools::contain(char* source, char character)
{
    int size = strlen(source);
    for (int i = 0; i < size ; i++)
    {
        if (source[i] == character)
            return true;
    }
    return false;
}

用法：

 std::cout << "postFix : " << Tools::toPostfix("a+(b*c+t)") << std::endl;
 std::cout << "Answer : " << Tools::calculatePostfix("24*95+-") << std::endl;
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