C++的RPN計算器

Question

我正在嘗試爲RPN計算器編寫代碼，我正在努力使它正確計算多個「左」和「右」操作數。

到目前爲止，我的代碼工作的投入，如： 10 15 + = 25

我無法搞清楚如何獲得這些輸入正確輸出：

輸入：100 10 50 25/* - -2/=輸出：-40

此外，現在我只是將堆棧大小除以2，它適用於某些錯誤檢查，但然後它打印錯誤的類似上/下輸入。 100 10 50 25/* - -2/= -40

我該如何獲得代碼來檢查下面這兩個輸入的錯誤？

輸入：10 20 */=輸出：錯誤：有太多的運營商

12 20 30/=     Error: Too many operand 

任何幫助將不勝感激，謝謝！

#include <iostream> 
#include <stack> 
#include <string> 
#include <sstream> //make use of a class called istringstream 
#include<iomanip> 

using namespace std; 

//Function prototype for isOperator 
bool isOperator(const string& input); 

//Function prototype for perforOperation 
int performOperation(const string& input, stack<double>& calcStack); 

int main() 
{ 

    cout << "RPN Calculator: " << endl; 
    cout << "Input\n"; 

    stack<double> calcStack; 
    string input; 


     while(input != "0") 
     { 
      //Terminate program when 0 is entered by user 
      while(input != "=") 
      { 

      // get input 
      cin >> input; 

      // check for being numeric value 
      double num; 

       if(istringstream(input) >> num) 
       { 
        //use push function 
        calcStack.push(num); 
       } 

       // check for operator 
       else if(isOperator(input)) 
       { 
        performOperation(input, calcStack); 
       } 

       // If user enters 0 on a line followed by a new line, the program exits  ???????????? 
       else if(input == "0\n") 
       { 
        return -1; 
       } 

       // invalid output check 
       //else 
       //{ 
        //cout << "Invalid input" << endl; 
       //} 
      } 
      } 
} 

    bool isOperator(const string& input) 
    { 
     string operators[] = {"-", "+", "*", "/"}; 

     for(int i=0; i<4; i++) 
     { 
      if(input == operators[i]) 
      { 
       return true; 
      } 
     } 

     return false; 
} 


int performOperation(const string& input, stack<double>& calcStack) 
{ 
    double firstOperand; 
    double secondOperand; 
    double result; 

    if(calcStack.size() > 2)      //Error check gives a false error for last input ??? 
    { 
     cout << "Error: too many operands" << endl; 
     return 1; 
    } 

    //Error chceck for too many operators   ////STILL PRINT OUTPUT??? 
    if(calcStack.size() < 2) 
     { 
      cout << "Error: too many operators" << endl; 
      return 1; 
     } 


    secondOperand = calcStack.top(); 
    calcStack.pop(); 


    firstOperand = calcStack.top(); 
    calcStack.pop(); 


    if(input == "-") 
    { 
     result = firstOperand-secondOperand; 
    } 

    else if (input == "+") 
    { 
     result = firstOperand + secondOperand; 
    } 

    else if (input == "*") 
    { 
     result = firstOperand * secondOperand; 
    } 

    else if(input == "/") 
    { 
    result = firstOperand/secondOperand; 
    } 


    // If user enters 0 on a line followed by a new line, the program exits   ??????????? 
    else if(input == "0\n") 
    { 
    return -1; 
    } 

     //Division by zero error 
     if(secondOperand == 0) 
      { 
       cout << "Error: Division by 0.\n"; 
       return -1; 
      } 

    cout << "Output\n"; 
    cout << result << endl; 
    calcStack.push(result); 

return 0; 

} 

Answer 1

這段代碼在這裏

if(calcStack.size() > 2)      
{ 
    cout << "Error: too many operands" << endl; 
    return 1; 
} 

需要移動到主稍有變換

else if(isOperator(input)) 
{ 
    performOperation(input, calcStack); 
} 
else if(input == "=") 
{ 
    if (calcStack.size() != 1) 
    { 
     cout << "Error: too many operands" << endl; 
     return 1; 
    } 
    else 
    { 
     cout << "Result: " << calcStack.top(); 
     // now decide whether or not you are preserving the result for 
     // the next computation 
     calcStack.pop(); // Assuming not keeping result 
    } 
} 

這意味着你將需要重新考慮這個循環while(input != "=")

你真的很親密。

兩條建議：

您可以優化isOperator函數。

bool isOperator(const string& input) 
{ 
    static const string operators ="-+*/"; 
    if (input.length() == 1) // right size to be an operator. 
    { 
     return operators.find_first_of(input[0]) != string::npos; 
     // look in the operator string for the first (and only) character in input 
    } 
    return false; 
} 

而且因爲你知道有在運營商只有一個字符，你可以使用一些更優雅比的if/else若：

switch (input[0]) 
{ 
    case '-': 
     result = firstOperand - secondOperand; 
     break; 
    case '+': 
     result = firstOperand + secondOperand; 
     break; 
    case '*': 
     result = firstOperand * secondOperand; 
     break; 
    case '/': 
     if (secondOperand == 0) 
     { // moved this test to here because it's the only place it matters. 
      cout << "Error: Division by 0.\n"; 
      return -1; 
     } 
     result = firstOperand/secondOperand; 
     break; 
} 

編對付評論

通過這種指出你的代碼應該已經改變了很多，你可能想用你當前的代碼開始一個新的問題。如果不是，這就是我在評論中所談論的。

else if(isOperator(input)) 
{ 
    if (performOperation(input, calcStack) != 0) 
    { 
     // empty out the stack and delete all pending user input. 
     calcStack.clear(); 
     cin.ignore(numeric_limits<streamsize>::max(), '\n'); 
     cout << "Bad input. Try again." << endl; 
     break; // exit input loop. 
    } 
} 
else if(input == "=") 
{ 
...