自上而下的解析器希望在“代码”中有像样的左递归案例。
大家好，Stack over Flow 成员。
我正在学习编译器课程。 我确实理解自顶向下解析器应该避免左递归，而转变为右递归方式。
问题是，
a）我是否正确理解自上而下解析器等于 LL 和自下而上解析器等于 LR ？
b) 我发现左递归是一个自称为 ex) Expr :== Expr '+' Term | 的规则可能导致无限循环查找 Expr 的项。但无论如何，有考虑用 C 或 Java 输入的示例代码吗？ （我不需要解析器或扫描器代码）我需要的是带有句子形式的案例代码示例，通过左递归发生无限循环。
c) 在自顶向下解析器中使用右递归实际上有什么区别？
ANS c) 消除了回溯的需要。但还有别的事吗？
ANS b) x - 2 * y 但还有其他什么？因为这个采用回溯解析方式。
我发现了非左递归和左递归的例子。
左递归语法
A -> Ax
非左递归语法
A -> Bx
B -> Ay
两者都陷入无限循环。
谢谢您并感谢您所有的专家。
Hello fellow stack over flow members.
I'm studying for compiler class.
I did understand Top-Down Parser should avoid left-recursion, and transform into right-recursion way.
Questions are,
a) am I understanding right Top-Down Parser is equal to LL and Bottom-Up Parser is equal to LR ?
b) I've found out left-recursion is Rule that calls itself ex) Expr :== Expr '+' Term | Term which can cause infinite loop to find Expr. But anyhow, any example code of consider input in C or Java? ( I don't want the parser or scanner code ) what I need is case code example with sentential form that occur infinite loop by left recursion.
c) What actually makes difference in a way using Right Recursion in Top-Down Parser?
ANS c) Eliminating the need to backtrack. but something else?
ANS b) x - 2 * y but also something else? because this one works with backtrack way of parsing.
Case example that I have found out the both non-left recursion and left recursion.
Left Recursion Grammar
A -> Ax
Non-Left Recursion Grammar
A -> Bx
B -> Ay
Both are getting into infinite loop.
Thank you and appreciated for all your expert.
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a）我是否理解正确自上而下解析器等于 LL 和自下而上解析器等于 LR ？
是的，
自上而下的解析器会进入左递归的无限循环，因为产生式在代码中看起来像这样：
A 永远调用自己，并且永远不会从输入流中删除任何内容。
左递归不一定是立即的，所以你的“非左递归语法”仍然是左递归的：
如果你用它的产生式替换 B，你可以看到它是左递归的：
这是一个正确转换左递归语法的示例到右递归语法：
左递归：
右递归：
E -> TB以来，规则E中-> E + T |完成规则应用后，T、T 将始终出现在最左侧。由于最左侧由 E -> 处理。 TB，我们可以自由地构造我们用 B -> 所做的字符串的右侧。 + T B. 我们需要一个 lambda 产生式来为我们提供递归的停止点。
A->斧头和A-> xA 是等价语法，只有一个是左递归的，另一个是右递归的。
a) am I understanding right Top-Down Parser is equal to LL and Bottom-Up Parser is equal to LR ?
yes
Top down parsers get into an infinite loop with left-recursion since the productions, in code, look like:
A calls itself forever and never removes anything from the input stream.
Left recursion doesn't have to be immediate so your "non left recursive grammar" is still left recursive:
You can see that it is left recursive if you replace B by its production:
Here is an example of correctly converting a left-recursive grammar to a right-recursive grammar:
Left recursive:
Right recursive:
E -> T B since, in the rule E -> E + T | T, T will ALWAYS appear on the left most side after finishing the application of the rule. Since the left most side is taken care of by E -> T B, we are free to construct the right side of the string which we do with B -> + T B. We need a lambda production to give us a stopping point for the recursion.
A -> Ax and A -> xA would be equivalent grammars, just one is left and the other is right recursive.