left-recursion

Left Factoring & Removing Left Recursion JavaCC

∥☆過路亽.° 提交于 2019-12-06 04:57:34
I have a grammar which I have to use JJTree and JavaCC to create a symbol table and an AST. While I fully understand the sections of my assignment to create the table and tree, the grammar I was given is ambiguous, contains left recursion and indirect left recusion. It also needs to be left factored. I have trawled all over the internet to try find methods that would work for me. For example: A ::= Aα | β can be changed to: A ::= βA' A' ::= αA' | ε But I don't know how to apply this to my grammar. Here is a section of the production rules I wrote from the grammar that contains the problems

Removing left recursion in DCG - Prolog

自闭症网瘾萝莉.ら 提交于 2019-12-06 02:14:01
问题 I've got a small problem with left recursion in this grammar. I'm trying to write it in Prolog, but I don't know how to remove left recursion. <expression> -> <simple_expression> <simple_expression> -> <simple_expression> <binary_operator> <simple_expression> <simple_expression> -> <function> <function> -> <function> <atom> <function> -> <atom> <atom> -> <number> | <variable> <binary_operator> -> + | - | * | / expression(Expr) --> simple_expression(SExpr), { Expr = SExpr }. simple_expression

Step by step elimination of this indirect left recursion

我怕爱的太早我们不能终老 提交于 2019-12-02 23:28:26
I've seen this algorithm one should be able to use to remove all left recursion. Yet I'm running into problems with this particular grammar: A -> Cd B -> Ce C -> A | B | f Whatever I try I end up in loops or with a grammar that is still indirect left recursive. What are the steps to properly implement this algorithm on this grammar? Rule is that you first establish some kind of order for non-terminals, and then find all paths where indirect recursion happens. In this case order would be A < B < C, and possible paths for recursion of non-terminal C would be C=> A => Cd and C=> B => Ce so new

Help with left factoring a grammar to remove left recursion

匆匆过客 提交于 2019-12-01 04:01:49
I have a small custom scripting language, and I am trying to update it to allow boolean expressions such as a > 2 and a > 2 and (b < 3 or c > 5) . It's the parenthetical expressions that I am having trouble with here. Here is a (edited since the original post based on the answer from @Bart Kiers) full grammar that exhibits the problem. This is a pared-down version of my actual grammar, but the problem occurs here too. grammar test; options { language = 'JavaScript'; output = AST; } statement : value_assignment_statement EOF ; value_assignment_statement : IDENT '=' expression ; value_expression

Help with left factoring a grammar to remove left recursion

倖福魔咒の 提交于 2019-12-01 01:57:23
问题 I have a small custom scripting language, and I am trying to update it to allow boolean expressions such as a > 2 and a > 2 and (b < 3 or c > 5) . It's the parenthetical expressions that I am having trouble with here. Here is a (edited since the original post based on the answer from @Bart Kiers) full grammar that exhibits the problem. This is a pared-down version of my actual grammar, but the problem occurs here too. grammar test; options { language = 'JavaScript'; output = AST; } statement

Why can't a LL grammar be left-recursive?

主宰稳场 提交于 2019-11-29 11:21:45
问题 In the dragon book , LL grammar is defined as follows: A grammar is LL if and only if for any production A -> a|b , the following two conditions apply. FIRST(a) and FIRST(b) are disjoint. This implies that they cannot both derive EMPTY If b can derive EMPTY , then a cannot derive any string that begins with FOLLOW(A) , that is FIRST(a) and FOLLOW(A) must be disjoint. And I know that LL grammar can't be left recursive, but what is the formal reason? I guess left-recursive grammar will

Recursive definitions with scala-parser-combinators

大兔子大兔子 提交于 2019-11-28 05:33:44
问题 I have been trying to build a SQL parser with the scala-parser-combinator library, which I've simplified greatly into the code below. class Expression case class FalseExpr() extends Expression case class TrueExpr() extends Expression case class AndExpression(expr1: Expression, expr2: Expression) extends Expression object SimpleSqlParser { def parse(sql: String): Try[Expression] = new SimpleSqlParser().parse(sql) } class SimpleSqlParser extends RegexParsers { def parse(sql: String): Try[_ <:

Remove Ambiguity in abstract syntax in other to write DCG parser Prolog

大城市里の小女人 提交于 2019-11-28 04:27:35
问题 P => Program K => Block S => Single-command C => Commands E => Expression B => Boolean-expr I => Identifier N > Numeral P ::= K. K ::= begin C end C ::= C1 ; C2 | S S ::= I := E | if (B) then S | if (B) then S1 else S2 | while (B) do S | repeat C until (B) | K | print E E ::= − E | E1 + E2 | E1 − E2 | E1 E2 | E1 div E2 | E1 mod E2 | (E) | I | N B ::= E1 = E2 | E1 > E2 | E1 < E2 | E1 != E2 | not B | B1 and B2 | B1 or B2 | (B) I am supposed to remove ambiguities in E and B so that I can write a