C# 解析缺少乘法符号的数学表达式

从年开始，我现在想写一个解析器的实现

// Handles * and / 
private static void Summand(Scanner scanner, ref TermNode currentTree, ref Token currentToken)
{
    Factor(scanner, ref currentTree, ref currentToken);

    while (currentToken is OperatorToken && _multiplicationOperators.Contains(((OperatorToken)currentToken).OperatorChar)) // So long as the token is * or /
    {
        TermNode node = new TermNode(currentTree, null, currentToken);
        currentTree = null;
        scanner.MoveNext();
        currentToken = scanner.Current;
        Factor(scanner, ref currentTree, ref currentToken); // Handles ^
        node.RightChild = currentTree;
        currentTree = node;
    }
}

// I think this one might be wrong...
private static void Factor(Scanner scanner, ref TermNode currentTree, ref Token currentToken)
{
    Exponent(scanner, ref currentTree, ref currentToken);

    while (currentToken is OperatorToken && ((OperatorToken)currentToken).OperatorChar == '^') // So long as the token is ^
    {
        TermNode node = new TermNode(currentTree, null, currentToken);
        currentTree = null;
        scanner.MoveNext();
        currentToken = scanner.Current;
        Exponent(scanner, ref currentTree, ref currentToken);
        node.RightChild = currentTree;
        currentTree = node;
    }
}

private static void Exponent(Scanner scanner, ref TermNode currentTree, ref Token currentToken)
{
    if (currentToken is BracketToken)
    {
        BracketToken bracketToken = (BracketToken)currentToken;
        if (bracketToken.IsOpening)
        {
            scanner.MoveNext();
            currentToken = scanner.Current;
            Term(scanner, ref currentTree, ref currentToken);
            if (currentToken is BracketToken && !((BracketToken)currentToken).IsOpening)
            {
                scanner.MoveNext();
                currentToken = scanner.Current;
            }
            else
                throw new ParseException("Unbalanced brackets!");
        }
    }
    else
    {
        node = new TermNode(null, null, currentToken);

        if (currentToken is OperatorToken)
        {
            currentTree = null;
            scanner.MoveNext();
            currentToken = scanner.Current;
            Exponent(scanner, ref currentTree, ref currentToken, false);
            node.RightChild = currentTree;
            currentTree = node;
        }
        else if (currentToken is VariableToken || currentToken is ConstantToken)
        {
            currentTree = node;
            scanner.MoveNext();
            currentToken = scanner.Current;
        }
        else if (currentToken is FunctionToken)
        {
            currentTree = null;
            scanner.MoveNext();
            currentToken = scanner.Current;
            Exponent(scanner, ref currentTree, ref currentToken, false);
            node.RightChild = currentTree;
            currentTree = node;
        }
    }

}

---

现在我想知道如何更改此方法以允许表达式，例如

3（a+b）

。。。以及这种方法和函数是否是正确的归档方式。

这只是一个修改（未声明的）语法的案例：

Summand = Factor | Summand "*" Factor | Summand "/" Factor ;

到

并根据需要修改手写的递归下降parrser

所以您需要修改“Summand”，这样它就不会检查显式乘法运算符，而是继续检查除法运算符

所以代码看起来像这样：

private static void Summand(Scanner scanner, ref TermNode currentTree, ref Token currentToken)
{
   Factor(scanner, ref currentTree, ref currentToken);
   while (true) // repeat for each implicit multiply or explicit divide
   {
     if (currentToken is OperatorToken && ((OperatorToken)currentToken).OperatorChar == '/')
     { // handle divide
       TermNode node = new TermNode(currentTree, null, currentToken);
       currentTree = null;
       scanner.MoveNext();
       currentToken = scanner.Current;
       Factor(scanner, ref currentTree, ref currentToken);
       node.RightChild = currentTree;
       currentTree = node;
     }
     else { // handle possible multiplication
            TermNode multiplicand = node ;
            Factor(scanner, ref currentTree, ref currentToken)
            if (Factor_failed) return; // no implicit product
            currentTree = new TermNode(multiplicand, currentTree,  maketoken("*"));              

          }
   } //while
} // Summand

您的解析器缺少的是来自每个子解析器的一个信号，该信号指示子解析器无法找到请求它解析的内容。（您需要实现“Factor_failed”的想法。）这与它发现的证据不同，证据表明请求解析的内容存在，但语法无效。我建议您将每个返回类型更改为“bool”，如果子parser成功，则返回“true”，如果它找不到应该解析的内容的证据，则返回“false”，如果它在解析中途失败，则抛出异常

---

查看一个实现这些想法的方法。

这怎么不是你3小时前提出并链接到此处的问题的副本@MarcinJuraszek这次我要的是C#。你能发布factor方法的代码吗。@Gaber ber我已经添加了我已经计算出的其余部分。你真的是说

factor（scanner，ref currentTree，ref currentToken）；TermNode节点=新的TermNode（currentTree，null，maketoken（“*”））；node.RightChild=currentTree？看起来好像“currentTree”被添加到节点两次…我试图让“*”分支使用“乘法”操作符构建一棵树，就像divide分支一样。。。啊，我想我明白了问题所在。我已经找到了答案。
private static void Summand(Scanner scanner, ref TermNode currentTree, ref Token currentToken)
{
   Factor(scanner, ref currentTree, ref currentToken);
   while (true) // repeat for each implicit multiply or explicit divide
   {
     if (currentToken is OperatorToken && ((OperatorToken)currentToken).OperatorChar == '/')
     { // handle divide
       TermNode node = new TermNode(currentTree, null, currentToken);
       currentTree = null;
       scanner.MoveNext();
       currentToken = scanner.Current;
       Factor(scanner, ref currentTree, ref currentToken);
       node.RightChild = currentTree;
       currentTree = node;
     }
     else { // handle possible multiplication
            TermNode multiplicand = node ;
            Factor(scanner, ref currentTree, ref currentToken)
            if (Factor_failed) return; // no implicit product
            currentTree = new TermNode(multiplicand, currentTree,  maketoken("*"));              

          }
   } //while
} // Summand