ll

Making a Grammar LL(1)

由 坚强是说给别人听的谎言 提交于 2019-12-22 04:57:10
问题 I have the following grammar: S → a S b S | b S a S | ε Since I'm trying to write a small compiler for it, I'd like to make it LL(1). I see that there seems to be a FIRST/FOLLOW conflict here, and I know I have to use substitution to resolve it, but I'm not exactly sure how to go about it. Here is my proposed grammar, but I'm not sure if it's correct: S-> aSbT | epsilon T-> bFaF| epsilon F-> epsilon Can someone help out? 回答1: In his original paper on LR parsing, Knuth gives the following

How to prove left-recursive grammar is not in LL(1) using parsing table

由 血红的双手。 提交于 2019-12-14 00:32:48
问题 I have a grammar and would like to prove that it is not in LL(1): S->SA|A A->a As it is a left-recursive grammarm, to find the first and follow sets I eliminated the left recursion and got: S->AS' S'->AS'|Empty A->a first of A={a} follow of S={$} first of s'={a,ε} follow of S'={$} first of S={a} follow of A={a,$} But when I filled in the parsing table, I did not get any cell with 2 entries. Then how is one to prove that the given grammar is not in LL(1)? 回答1: First of all you are finding

Finding a language that is not LL(1)?

由 亡梦爱人 提交于 2019-12-13 13:05:59
问题 I've been playing around with a lot of grammars that are not LL(1) recently, and many of them can be transformed into grammars that are LL(1). However, I have never seen an example of an unambiguous language that is not LL(1). In other words, a language for which any unambiguous grammar for the language is not LL(1)), nor do I have any idea how I would go about proving that I had found one if I accidentally stumbled across one. Does anyone know how to prove that a particular unambiguous

Ebnf – Is this an LL(1) grammar?

由 会有一股神秘感。 提交于 2019-12-12 12:27:07
问题 I found the following EBNF on wikipedia, describing EBNF: letter = "A" | "B" | "C" | "D" | "E" | "F" | "G" | "H" | "I" | "J" | "K" | "L" | "M" | "N" | "O" | "P" | "Q" | "R" | "S" | "T" | "U" | "V" | "W" | "X" | "Y" | "Z" ; digit = "0" | "1" | "2" | "3" | "4" | "5" | "6" | "7" | "8" | "9" ; symbol = "[" | "]" | "{" | "}" | "(" | ")" | "<" | ">" | "'" | '"' | "=" | "|" | "." | "," | ";" ; character = letter | digit | symbol | "_" ; identifier = letter , { letter | digit | "_" } ; terminal = "'"

Can a ANTLR grammar file be modified to be used by PLY?

由 北战南征 提交于 2019-12-10 11:44:58
问题 I want to make a python program that uses PLY to parse Javascript files, I din't found any sources of parsers that implement the ECMAScript, Javascript rules that use PLY. The only thing I found was some ANTLR grammar files to parse javascript and ecmascript: http://www.antlr.org/grammar/1153976512034/ecmascriptA3.g http://www.antlr.org/grammar/1206736738015/JavaScript.g Can ANTLR grammar files be adapted to be used as PLY rules, if yes, how can be done in a semi-automatic way, do I need to

How to manually construct an AST?

由 牧云@^-^@ 提交于 2019-12-05 20:45:59
问题 I'm currently learning about parsing but i'm a bit confused as how to generate an AST. I have written a parser that correctly verifies whether an expressions conforms to a grammar (it is silent when the expression conforms and raises an exception when it is not). Where do i go from here to build an AST? I found plenty of information on building my LL(1) parser, but very little on then going on to build the AST. My current code (written in very simple Ruby, and including a lexer and a parser)

Relationship between LR(0), LL(0), LALR(1), etc?

由 有些话、适合烂在心里 提交于 2019-12-05 15:49:31
I'm really struggling to unterstand the relationship between: LR(0) LL(0) LALR(1) SLR(1) LR(1) LL(1) I'm pretty sure LALR(1) and SLR(1) are subsets of LR(1), but I'm lost about the others. Are they all exclusive? Is LL(0) a subset of LL(1)? Thanks The containment rules are the following: Every LR(0) grammar is also SLR(1), but not all SLR(1) grammars are LR(0). Every SLR(1) grammar is also LALR(1), but not all LALR(1) grammars are SLR(1). Every LALR(1) grammar is also LR(1), but not all LR(1) grammars are LALR(1). Every LL(1) grammar is also LR(1), but not all LR(1) grammars are LL(1). Every

Making a Grammar LL(1)

由 自闭症网瘾萝莉.ら 提交于 2019-12-05 04:46:41
I have the following grammar: S → a S b S | b S a S | ε Since I'm trying to write a small compiler for it, I'd like to make it LL(1). I see that there seems to be a FIRST/FOLLOW conflict here, and I know I have to use substitution to resolve it, but I'm not exactly sure how to go about it. Here is my proposed grammar, but I'm not sure if it's correct: S-> aSbT | epsilon T-> bFaF| epsilon F-> epsilon Can someone help out? templatetypedef In his original paper on LR parsing , Knuth gives the following grammar for this language, which he conjectures "is the briefest possible unambiguous grammar

Performance of parsers: PEG vs LALR(1) or LL(k)

由 心不动则不痛 提交于 2019-12-03 16:40:49
问题 I've seen some claims that optimized PEG parsers in general cannot be faster than optimized LALR(1) or LL(k) parsers. (Of course, performance of parsing would depend on a particular grammar.) I'd like to know if there are any specific limitations of PEG parsers, either valid in general or for some subsets of PEG grammars that would make them inferior to LALR(1) or LL(k) performance-wise. In particular, I'm interested in parser generators, but assume that their output can be tweaked for

Example of an LR grammar that cannot be represented by LL?

由 痞子三分冷 提交于 2019-12-03 10:09:47
All LL grammars are LR grammars, but not the other way around, but I still struggle to deal with the distinction. I'm curious about small examples, if any exist, of LR grammars which do not have an equivalent LL representation. Chris Dodd Well, as far as grammars are concerned, its easy -- any simple left-recursive grammar is LR (probably LR(1)) and not LL. So a list grammar like: list ::= list ',' element | element is LR(1) (assuming the production for element is) but not LL. Such grammars can be fairly easily converted into LL grammars by left-factoring and such, so this is not too
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