关于表达式中除数为0的容错机制的处理

说明：对于一个表达式（比如1+2/(A-B)），如果A-B为0，有时我们想这个表达式返回1，而不是0或者抛出异常。也就是局部的除数异常是允许的。这个时候就需要对公式做容错处理。

下面是用C#的处理过程，如有错误，欢迎指正

/// <summary>/// 公式容错/// 在执行公式时，会有除数为0的情况，此时不应该报错，只是包含除数为0的表达式计算结果为0，整个表达式继续执行/// 如1+(2/(4-8/(1+1)))  执行结果应该是1 而不应该报错/// 方法：/// 1：将原表达式转换为后缀表达式/// 2：再将后缀表达式转换为容错之后的中缀表达式/// 3：栈顶元素即为结果/// </summary>public class FormulasConvert {//存原始后缀表达式Stack<char> _suffixStack = new Stack<char>();/// <summary>/// 后缀表达式结果/// </summary>public string _suffixFormula = string.Empty;//存容错后中缀表达式Stack<string> _infixingStack = new Stack<string>();/// <summary>/// 容错后中缀表达式结果/// </summary>public string _infixingFormula = string.Empty;/// <summary>/// 原始公式/// </summary>private string _inputFormula = string.Empty;/// <summary>/// 用字符转换之后的原始公式/// 10+10==》A+B/// </summary>private string _newFormula = string.Empty;//操作符private string _operator = "+-*/()";/// <summary>/// 原公式：100+2/300/// 对应：/// A:100  B:2 C:300/// 最终/// A+B/C/// </summary>Dictionary<string, string> _dicCompary = new Dictionary<string, string>();public FormulasConvert(string inputFormula) {this._inputFormula = inputFormula;}/// <summary>/// 将原始表达式转换成后缀表达式/// </summary>/// <param name="inputString">原中缀表达式</param>/// <returns></returns>public string ConvertToSuffix() {OperatorFormula();for (int i = 0; i < _newFormula.Length; i++) {char ch = _newFormula[i];if (!IsOperator(ch)) {_suffixFormula += ch;} else if (ch == '(')_suffixStack.Push(ch);else if (ch == ')') {while (_suffixStack.Peek() != '(')_suffixFormula += _suffixStack.Pop();_suffixStack.Pop();} else if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {if (_suffixStack.Count == 0)_suffixStack.Push(ch);else if (_suffixStack.Peek() == '(')_suffixStack.Push(ch);else if (CompareOperate(ch, _suffixStack.Peek()) == 1)_suffixStack.Push(ch);else if (CompareOperate(ch, _suffixStack.Peek()) == 0) {_suffixFormula += _suffixStack.Pop();_suffixStack.Push(ch);} else if (CompareOperate(ch, _suffixStack.Peek()) == -1) {int com = -1;while (com == -1 || com == 0) {_suffixFormula += _suffixStack.Pop();if (_suffixStack.Count != 0)com = CompareOperate(ch, _suffixStack.Peek());elsebreak;}_suffixStack.Push(ch);}}}string tmpStr = string.Empty;for (int i = 0; i < _suffixStack.Count + 1; i++) {if (_suffixStack.Count > 0) {tmpStr += _suffixStack.Peek().ToString();_suffixStack.Pop();}}_suffixFormula = _suffixFormula + tmpStr;return GetFormulaResult(_suffixFormula);}private int CompareOperate(char ch, char stackCh) {if (ch == stackCh)return 0;else if ((ch == '+' && stackCh == '-') || (ch == '-' && stackCh == '+'))//表示等于return 0;else if ((ch == '*' && stackCh == '/') || (ch == '/' && stackCh == '*'))return 0;else if ((ch == '+' || ch == '-') && (stackCh == '*' || stackCh == '/'))//表示小于return -1;else if ((ch == '*' || ch == '/') && (stackCh == '+' || stackCh == '-'))//返回1 表示输入运算符的优先级大于栈顶运算符return 1;elsereturn -2;}/// <summary>/// 将后缀表达式生成中缀表达式（未容错）/// </summary>/// <returns></returns>public string ConvertToInfixing() {if (string.IsNullOrEmpty(this._inputFormula))return string.Empty;ConvertToSuffix();if (string.IsNullOrEmpty(this._suffixFormula))return string.Empty;for (int i = 0; i < _suffixFormula.Length; i++) {char ch = _suffixFormula[i];if (!IsOperator(ch)) { //操作数入栈_infixingStack.Push(ch.ToString());} else if (ch == '+' || ch == '-' || ch == '*' || ch == '/') {string tmp1 = _infixingStack.Pop();string tmp2 = _infixingStack.Pop();string tmp = string.Empty;tmp = "(" + tmp2;tmp += ch;tmp += tmp1 + ")";_infixingStack.Push(tmp);}}return GetFormulaResult(_infixingStack.Peek());}/// <summary>/// 将后缀表达式生成中缀表达式（容错,替换除数为0）/// 返回容错之后的新中缀表达式/// </summary>/// <param name="inputString"></param>/// <returns></returns>public string Convert() {if (string.IsNullOrEmpty(this._inputFormula))return string.Empty;ConvertToSuffix();if (string.IsNullOrEmpty(this._suffixFormula))return string.Empty;for (int i = 0; i < _suffixFormula.Length; i++) {char ch = _suffixFormula[i];if (!IsOperator(ch)) { //操作数入栈_infixingStack.Push(ch.ToString());} else if (ch == '+' || ch == '-' || ch == '*') {string tmp1 = _infixingStack.Pop();string tmp2 = _infixingStack.Pop();string tmp = string.Empty;tmp = "(" + tmp2;tmp += ch;tmp += tmp1 + ")";_infixingStack.Push(tmp);} else if (ch == '/') {//(A if(Ture) else B)string tmp1 = _infixingStack.Pop();string tmp2 = _infixingStack.Pop();string tmp = string.Empty;tmp = "(0 if(" + tmp1 + "==0) else (" + tmp2 + "/" + tmp1 + "))";_infixingStack.Push(tmp);}}return GetFormulaResult(_infixingStack.Peek());}/// <summary>/// 处理原始表达式为原运算符/// 最终变成A+B/C这种模式/// </summary>private void OperatorFormula() {if (string.IsNullOrEmpty(this._inputFormula))return;//26个字母应该够用了，一个公式有26个运算项已经够多了char start = 'A';string tmpData = string.Empty;for (int i = 0; i < this._inputFormula.Length; i++) {char ch = _inputFormula[i];if (IsOperator(ch)) {if (!string.IsNullOrEmpty(tmpData.Trim())) {this._dicCompary.Add(start.ToString(), tmpData);_newFormula += start.ToString();start = (char)((int)start + 1);tmpData = string.Empty;} else {tmpData = string.Empty;  //空格不处理}_newFormula += ch.ToString();} else {tmpData += _inputFormula[i].ToString();}}if (!string.IsNullOrEmpty(tmpData.Trim())) {_dicCompary.Add(start.ToString(), tmpData);_newFormula += start.ToString();  //加上最后一个操作数}}/// <summary>/// 根据_newFormula与_dicCompary重新组织成公式/// </summary>/// <returns></returns>private string GetFormulaResult(string formulas) {if (string.IsNullOrEmpty(formulas))return string.Empty;string newFormulas = string.Empty;for (int i = 0; i < formulas.Length; i++) {string key = formulas[i].ToString();if (_dicCompary.ContainsKey(key)) {newFormulas += _dicCompary[key];} else {newFormulas += key;}}return newFormulas;}//判断是否是操作符，目前操作符为+-*/()private bool IsOperator(char c) {return _operator.Contains(c.ToString());}}

private void button2_Click(object sender, EventArgs e) {FormulasConvert convert = new FormulasConvert(this.label2.Text);richTextBox2.Text = convert.ConvertToSuffix();}private void button3_Click(object sender, EventArgs e) {FormulasConvert convert = new FormulasConvert(this.label2.Text);richTextBox3.Text = convert.Convert();}