C++中的後綴計算結果只在一些時間正確的答案

Question

我目前正在學習數據結構，我試圖用C++編寫一個程序，它使用一個數組作爲模擬堆棧。代碼應該像這樣工作：

用戶輸入一個Postfix表達式。表達式從左到右讀取。每次讀取數字時，它都會「彈出」到模擬堆棧上。當讀取操作數時，模擬堆棧中的前兩個數字將被「彈出」，操作數將使用這些數字執行計算，並將得到的答案「推回」到模擬堆棧上。該過程繼續，直到模擬堆棧中只剩下一個數字。這個數字是Postfix表達式的答案，它在屏幕上向用戶讀出。

我的代碼大部分工作在下面。但是，如果我模擬堆棧中的其中一個值大於或等於128，那麼該數字似乎存儲不正確（例如，+128將變成-128）。

我在做什麼錯？它與calc函數中的答案有什麼關係？具體來說，這行代碼：return answer;

任何幫助或提示將不勝感激。

#include <string> 
    #include <iostream> 
    using namespace std; 

    // Define the size of the stack: 
    #define SIZE 50 

    // Global variable declarations: 
    char stack[SIZE]; 
    int top = -1; // The top of the array 

    // Function for PUSHING operands onto the stack: 
    void push(char elem) 
    { 
     top++; 
     stack[top] = elem; 
    // cout << elem << " TESTING PUSH OUTPUT\n"; // TESTING 
    } 

    // Function for POPPING operands from the stack: 
    char pop() 
    { 
     char answer; 
     answer = stack[top]; 
     top--; 
    // cout << top << " TESTING WHAT'S POPPED\n"; // TESTING 
     return (answer); 
    } 

    // Function for CALCULATING two popped numbers: 
    double calc(char op) 
    { 
     int num1, num2, answer; 

     // Pop two numbers from the stack: 
     num1 = pop(); 

     num2 = pop(); // put the second popped number into 'num2' 

     // Calculate the two popped numbers: 
     switch (op) 
     { 
     case '*': 
      answer = num2 * num1; 
      break; 
     case '/': 
      answer = num2/num1; 
      break; 
     case '+': 
      answer = num2 + num1; 
      break; 
     case '-': 
      answer = num2 - num1; 
      break; 
     default: 
      cout << "Invalid expression!"; 
     } 

     // Push the result of the calculation back onto the stack: 
     push(answer); 

     return answer; 

    } 

    // The main program, which takes in the expression: 
    int main() 
    { 
     string postfix; 

     int solution; 

     cout << "Please enter a Postfix expression, with operators and operands separated by spaces and/or commas (e.g: 3 7 5 + 2 - * 16 4 + 10//): \n"; 
     getline(cin,postfix); 

     // If the User's Postfix expression begins with an operator, then throw up an error message: 
     if (postfix.find('/') == 0 || postfix.find('*') == 0 || postfix.find('+') == 0 || postfix.find('-') == 0) 
     { 
      cout << "Sorry, that's an invalid Postfix expression, and it can't be evaluated."; 

      return 0; 

     } 

     // FOR loop to read the Postfix expression: 
     for(unsigned int i = 0; i < postfix.length(); i++) 
     { 

      // If the value is a space or a comma, move on to the next value: 
      if(postfix[i] == ' ' || postfix[i] == ',') 
      { 
       continue; 
      } // End IF 

      // If the value is a number, extract it, and continue reading and extracting the next value until it is NOT a number: 
      else if(isalnum(postfix[i])) 
      { 

       int operand = 0; 

       while(i < postfix.length() && isalnum(postfix[i])) 
       { 
        // For multi-digit numbers, multiply the initial extracted number by 10 to move it into the 'tens' column, then add the next number to the initial number. 
        // Continue doing this until the next value is NOT a number 
        operand = (operand * 10) + (postfix[i] - '0'); 
        i++; 
       } // End WHILE 

       // When WHILE exits, 'i' will be non-numeric, of the 'end of the string'. Decrement 'i', because it will be incremented in the FOR loop as longs as the FOR condition is true 
       // (Stops 'i' from being incremented twice) 
       i--; 

       // Push the operand onto the stack: 
       push(operand); 
      } // End ELSE IF 

      // If the value is an operator, run the calculation function: 
      else 
      { 
       calc(postfix[i]); 
      } // End ELSE 

     } // End FOR 

     solution = pop(); // put the solution of the equation into 'answer' 

     cout << "The solution to the Postfix expression is: " << solution; 

    } 

Answer 1

operand類型爲int而push預計爲char。

因爲在大多數機器sizeof(int) > sizeof(char)上，如果operand的值試圖存儲在像char這樣的較小數據類型中，它的值可能會被截斷。

例如，我們假設sizeof(int) == 4和sizeof(char) == 1，一個字節是8位。現在，如果只設置了位（從0開始），則char將不能保存該信息，並且值0x100被截斷爲0x00。

爲了解決這個問題，可以使堆棧處理ints或將operand定義爲char。

Answer 2

所有你需要做的是讓你的模擬int類型，而不是char（和其他相關的變化）的stack。

Answer 3

你的棧是char類型，通常被限制在-128到127之間。嘗試unsigned char，它可能是0到255.如果你的棧上需要更大的值，那麼你將使它成爲一個整數或更大的東西。