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I've been reading on both in Wikipedia, and noticed that although LR(0) parsers exist, there's no such thing as LL(0) parser. From what I read, I understand that the k in LL(k)/LR(k) means how many characters the parser can see beyond the current character that it's currently working on. So my question is, why is there no such thing as LL(0) parser even though LR(0) exists? The difference has to do with what the k means in LR(k) versus LL(k). In LL(k), the parser maintains information about a top-down, left-to-right parse that traces out a leftmost derivation. The parser works by repeatedly

What advantages do LL parsers have over LR parsers to warrant their relative popularity in today's parser generator tools ? According to Wikipedia , LR parsing appears to have advantages over LL: LR parsing can handle a larger range of languages than LL parsing, and is also better at error reporting, i.e. it detects syntactic errors when the input does not conform to the grammar as soon as possible. This is in contrast to an LL(k) (or even worse, an LL(*) parser) which may defer error detection to a different branch of the grammar due to backtracking, often making errors harder to localize

So I've been trying to write a calculator with Scala's parser, and it's been fun, except that I found that operator associativity is backwards, and that when I try to make my grammar left-recursive, even though it's completely unambiguous, I get a stack overflow. To clarify, if I have a rule like: def subtract: Parser[Int] = num ~ "-" ~ add { x => x._1._1 - x._2 } then evaluating 7 - 4 - 3 comes out to be 6 instead of 0. The way I have actually implemented this is that I am composing a binary tree where operators are non-leaf nodes, and leaf nodes are numbers. The way I evaluate the tree is

I have a left recursive issue in my Antlr grammar. While I think I understand why there is a problem I am unable to think of a solution. The issue is with the last line for my datatype rule. I have included the entire grammar for you to see: grammar Test; options {output=AST;ASTLabelType=CommonTree;} tokens {FUNCTION; ATTRIBUTES; CHILDREN; COMPOSITE;} program : function ; function : ID (OPEN_BRACKET (attribute (COMMA? attribute)*)? CLOSE_BRACKET)? (OPEN_BRACE function* CLOSE_BRACE)? SEMICOLON? -> ^(FUNCTION ID ^(ATTRIBUTES attribute*) ^(CHILDREN function*)) ; attribute : ID (COLON | EQUALS)

Is there a good resource online with a collection of grammars for some of the major parsing algorithms (LL(1), LR(1), LR(0), LALR(1))? I've found many individual grammars that fall into these families, but I know of no good resource where someone has written up a large set of example grammars. Does anyone know of such a resource? Parsing Techniques - A Practical Guide has several examples (i.e. probably half a dozen or so per type) of almost every type of grammar. You can purchase the 2nd edition book, although the 1st edition is available for free on the author's website in PDF form (near

I've recently being trying to teach myself how parsers (for languages/context-free grammars) work, and most of it seems to be making sense, except for one thing. I'm focusing my attention in particular on LL(k) grammars , for which the two main algorithms seem to be the LL parser (using stack/parse table) and the Recursive Descent parser (simply using recursion). As far as I can see, the recursive descent algorithm works on all LL(k) grammars and possibly more, whereas an LL parser works on all LL(k) grammars. A recursive descent parser is clearly much simpler than an LL parser to implement,

Can anyone give me a simple example of LL parsing versus LR parsing? templatetypedef At a high level, the difference between LL parsing and LR parsing is that LL parsers begin at the start symbol and try to apply productions to arrive at the target string, whereas LR parsers begin at the target string and try to arrive back at the start symbol. An LL parse is a left-to-right, leftmost derivation. That is, we consider the input symbols from the left to the right and attempt to construct a leftmost derivation. This is done by beginning at the start symbol and repeatedly expanding out the