中缀表示法变为后缀表示法
但是我看不懂题目是要我们做什么  
谁能为我解释一下各个步骤分别是什么意思吗？？？  万分感谢呀！！  请大家帮帮忙
程序可以变精简些吗？？ 怎么改
#include <iostream>
#include <string>
#include <stack>
#include <cctype>  // to use the to lower function

using namespace std;


void Convert(const string & Infix, string & Postfix);

bool IsOperand(char ch);

bool TakesPrecedence(char OperatorA, char OperatorB);


int main(void)
  {
  char Reply;

  do
      {
      string Infix, Postfix;  // local to this loop

      cout < < "Enter an infix expression (e.g. (a+b)/c^2, with no spaces):"
        < < endl;
      cin >> Infix;

      Convert(Infix, Postfix);
      cout < < "The equivalent postfix expression is:" < < endl
        < < Postfix < < endl;
      cout < < endl < < "Do another (y/n)? ";
      cin >> Reply;
      }
  while (tolower(Reply) == 'y');

  return 0;
  }


/* Given:  ch  A character.
  Task:  To determine whether ch represents an operand (here understood
          to be a single letter or digit).
  Return: In the function name: true, if ch is an operand, false otherwise.
*/
bool IsOperand(char ch)
  {
  if (((ch >= 'a') && (ch <= 'z')) ||
      ((ch >= 'A') && (ch <= 'Z')) ||
      ((ch >= '0') && (ch <= '9')))
      return true;
  else
      return false;
  }


/* Given:  OperatorA    A character representing an operator or parenthesis.
          OperatorB    A character representing an operator or parenthesis.
  Task:  To determine whether OperatorA takes precedence over OperatorB.
  Return: In the function name: true, if OperatorA takes precedence over
          OperatorB.
*/
bool TakesPrecedence(char OperatorA, char OperatorB)
  {
  if (OperatorA == '(')
      return false;
  else if (OperatorB == '(')
      return false;
  else if (OperatorB == ')')
      return true;
  else if ((OperatorA == '^') && (OperatorB == '^'))
      return false;
  else if (OperatorA == '^')
      return true;
  else if (OperatorB == '^')
      return false;
  else if ((OperatorA == '*') || (OperatorA == '/'))
      return true;
  else if ((OperatorB == '*') || (OperatorB == '/'))
      return false;
  else
      return true;
      
  }

/* Given:  Infix    A string representing an infix expression (no spaces).
  Task:  To find the postfix equivalent of this expression.
  Return: Postfix  A string holding this postfix equivalent.
*/
void Convert(const string & Infix, string & Postfix)
  {
  stack <char> OperatorStack;
  char TopSymbol, Symbol;
  int k;

  for (k = 0; k < Infix.size(); k++)
      {
      Symbol = Infix[k];
      if (IsOperand(Symbol))
        Postfix = Postfix + Symbol;
      else
        {
        while ((! OperatorStack.empty()) &&
            (TakesPrecedence(OperatorStack.top(), Symbol)))
            {
            TopSymbol = OperatorStack.top();
            OperatorStack.pop();
            Postfix = Postfix + TopSymbol;
            }
        if ((! OperatorStack.empty()) && (Symbol == ')'))
            OperatorStack.pop();  // discard matching (
        else
            OperatorStack.push(Symbol);
        }
      }

  while (! OperatorStack.empty())
      {
      TopSymbol = OperatorStack.top();
      OperatorStack.pop();
      Postfix = Postfix + TopSymbol;
      }
  }