前些日子面試一個開發工作，考官出了這么一筆試題目，要我寫出實現過程， 思量半天，終于
用 C++ 完成，現將代碼貼出，與諸同道共分享。

// 頭文件 Calc.h
#ifndef __CALC_H__
#define __CALC_H__

#include <stack>

#define ascii_int(x) (x >= 0x30 && x <= 0x39) ? (x - 0x30) : (x)
const int GREATER =  1;
const int EQUAL   =  0;
const int LESS    = -1;

class Calculate {
public:
  int  evaluteExpr(char *exp);

private:
  int  getLevel(char ch);
  bool isOperator(char ch);
  int  compareOpteratorLevel(char inputChar, char optrStackTop);
  int  calc(int num1, int num2, char op);
  void evaluate(char ch);

private:
  std::stack<int>  _opnd_stack;
  std::stack<char> _optr_stack;
  static char _optr[];
  static int  _level[];
};

#endif


// 頭文件的實現代碼 Calc.cxx
#include "Calc.h"

char Calculate::_optr[] = {'#', '(', '+', '-', '*', '/', ')'};
int Calculate::_level[] = { 0,   1,   2,   2,   3,   3,   4 };

// Get current operator level for calculating
int Calculate::getLevel(char ch) {
  for (int i = 0; *(_optr+i) != '\0'; ++i)
    if (*(_optr+i) == ch)
      return *(_level+i);
}

// Calculate the operands
int Calculate::calc(int num1, int num2, char op) {
  switch (op)
    {
    case '+':
      return num1 + num2;
    case '-':
      return num1 - num2;
    case '*':
      return num1 * num2;
    case '/':
      return num1 / num2;
    }
}

// judge inputing character is operator or not
bool Calculate::isOperator(char ch) {
  for (char *p = _optr; *p != '\0'; ++p)
    if (*p == ch)
      return true;

  return false;
}

// Compare level of input operator and the top operator of operator stack
int Calculate::compareOpteratorLevel(char inputChar, char optrStackTop) {
//   if (inputChar == '(' && optrStackTop == ')')
//     return EQUAL;
//   else
  if (inputChar == '(')
    return GREATER;

  if (inputChar == ')' && optrStackTop == '(')
    return EQUAL;
  else if (inputChar == ')')
    return LESS;

  if (inputChar == '#' && optrStackTop == '#')
    return EQUAL;
//   else if (inputChar == '#')
//     return LESS;

  return (getLevel(inputChar) > getLevel(optrStackTop)) ? GREATER : LESS;
}

// Evaluate value while inputing operators
void Calculate::evaluate(char ch) {
  char op;
  int num, result;

  if (!isOperator(ch)) {
    _opnd_stack.push(ascii_int(ch));
    return ;
  }

  switch (compareOpteratorLevel(ch, _optr_stack.top()))
    {
    case GREATER :
      _optr_stack.push(ch);
      break;

    case EQUAL :
      _optr_stack.pop();
      break;

    case LESS :
      num = _opnd_stack.top();
      _opnd_stack.pop();

      result = _opnd_stack.top();
      _opnd_stack.pop();

      op = _optr_stack.top();
      _optr_stack.pop();

      result = calc(result, num, op);
      _opnd_stack.push(result);
      evaluate(ch);
      break;
    }
}

// Evaluate user specified expression
int Calculate::evaluteExpr(char *exp) {
  _optr_stack.push('#');
  for (char *p =exp; *p != '\0'; ++p )
    evaluate(*p);

  int result = _opnd_stack.top();
  _opnd_stack.pop();

  return result;
}


// 測試代碼 calc_test.cxx
#include <iostream>
#include "Calc.h"
using namespace std;

int main(void) {
  Calculate *calc = new Calculate();
  cout << "1+3*(4+7) = "
       << calc->evaluteExpr("1+3*(4+7)#")
       << endl;
  cout << "((1+2)) = "
       << calc->evaluteExpr("((1+2))#")
       << endl;
  cout << "3*8+9/7-5-9+(1-9)/4 = "
       << calc->evaluteExpr("3*8+9/7-5-9+(1-9)/4#")
       << endl;
  cout << "(6-7)*(5+9) = "
       << calc->evaluteExpr("(6-7)*(5+9)#")
       << endl;
  cout << "0*8+0/6-9+(7-1) = "
       << calc->evaluteExpr("0*8+0/6-9+(7-1)#")
       << endl;

  delete calc;
}

用 MinGW/G++ 3.4.5 編譯如下:
  g++  -o test.exe  Calc.cxx  Calc_test.cxx

作為一個演示算法夠了， 但代碼還是有一些缺點:
   (1) 只能處理一位數的加、減、乘、除表達式計算(可帶括號)
   (2) 沒有任何錯誤處理，例如不能在表達式中有空格