Question

最后，来自this question，问题仍然存在，这个subparser ......

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;
    }
}

...没有正确处理指数运算符（＆＃34; ^＆＃34;）。这是因为它是正确的关联。上面的代码处理它就好像它是左关联的。

例如：文字e^x^2被解释为(e^x)^2。但是，正确的解释＆＃34;将是e^(x^2)。

我已经尝试过这样的事情：

if (/* The current token is ^ */)
{
    TermNode node = new TermNode(tree, null, currentToken);
    tree = null;
    scanner.MoveNext();
    currentToken = scanner.Current;
    Exponent(ref tree);
    node.RightChild = tree;
    tree = node;
}
while (/* The current token is ^  */)
{
    TermNode detachedExponent = tree.RightChild;
    TermNode oldTree = tree;
    Token token = currentToken;
    tree.RightChild = null;
    tree = null;
    scanner.MoveNext();
    currentToken = scanner.Current;
    Exponent(ref tree);
    oldTree.RightChild = new TermNode(distachedExponent, tree, token);
    tree = oldTree;
}

仅适用于两个连续的＆＃34; ^＆＃34; - 表达式。不像e^x^y^z那样（e^(x^(y^z))而不是e^((x^y)^z)就像解析器声称的那样...我错过了什么？

Answer 1

如果您有a^b，并且看到^c，则会将其注入顶级^的RHS，并创建a^(b^c)，并留下结果充分表达。当您看到^d后，再次将其注入顶级^的RHS，创建a^((b^c)^d)。您不应该将其注入顶级^的RHS，而应注入右侧/最内侧^表达式。要实现这一点，只需在单独的变量中跟踪该表达式即可。然后，不要修改顶级表达式RightChild属性，而是修改子项。

数学表达式解析器

1 个答案: