Shift/Reduce conflicts in a propositional logic parser in Happy - parsing

I'm making a simple propositional logic parser on happy based on this BNF definition of the propositional logic grammar, this is my code
{
module FNC where
import Data.Char
import System.IO
}
-- Parser name, token types and error function name:
--
%name parse Prop
%tokentype { Token }
%error { parseError }
-- Token list:
%token
var { TokenVar $$ } -- alphabetic identifier
or { TokenOr }
and { TokenAnd }
'¬' { TokenNot }
"=>" { TokenImp } -- Implication
"<=>" { TokenDImp } --double implication
'(' { TokenOB } --open bracket
')' { TokenCB } --closing bracket
'.' {TokenEnd}
%left "<=>"
%left "=>"
%left or
%left and
%left '¬'
%left '(' ')'
%%
--Grammar
Prop :: {Sentence}
Prop : Sentence '.' {$1}
Sentence :: {Sentence}
Sentence : AtomSent {Atom $1}
| CompSent {Comp $1}
AtomSent :: {AtomSent}
AtomSent : var { Variable $1 }
CompSent :: {CompSent}
CompSent : '(' Sentence ')' { Bracket $2 }
| Sentence Connective Sentence {Bin $2 $1 $3}
| '¬' Sentence {Not $2}
Connective :: {Connective}
Connective : and {And}
| or {Or}
| "=>" {Imp}
| "<=>" {DImp}
{
--Error function
parseError :: [Token] -> a
parseError _ = error ("parseError: Syntax analysis error.\n")
--Data types to represent the grammar
data Sentence
= Atom AtomSent
| Comp CompSent
deriving Show
data AtomSent = Variable String deriving Show
data CompSent
= Bin Connective Sentence Sentence
| Not Sentence
| Bracket Sentence
deriving Show
data Connective
= And
| Or
| Imp
| DImp
deriving Show
--Data types for the tokens
data Token
= TokenVar String
| TokenOr
| TokenAnd
| TokenNot
| TokenImp
| TokenDImp
| TokenOB
| TokenCB
| TokenEnd
deriving Show
--Lexer
lexer :: String -> [Token]
lexer [] = [] -- cadena vacia
lexer (c:cs) -- cadena es un caracter, c, seguido de caracteres, cs.
| isSpace c = lexer cs
| isAlpha c = lexVar (c:cs)
| isSymbol c = lexSym (c:cs)
| c== '(' = TokenOB : lexer cs
| c== ')' = TokenCB : lexer cs
| c== '¬' = TokenNot : lexer cs --solved
| c== '.' = [TokenEnd]
| otherwise = error "lexer: Token invalido"
lexVar cs =
case span isAlpha cs of
("or",rest) -> TokenOr : lexer rest
("and",rest) -> TokenAnd : lexer rest
(var,rest) -> TokenVar var : lexer rest
lexSym cs =
case span isSymbol cs of
("=>",rest) -> TokenImp : lexer rest
("<=>",rest) -> TokenDImp : lexer rest
}
Now, I have two problems here
For some reason I get 4 shift/reduce conflicts, I don't really know where they might be since I thought the precedence would solve them (and I think I followed the BNF grammar correctly)...
(this is rather a Haskell problem) On my lexer function, for some reason I get parsing errors on the line where I say what to do with '¬', if I remove that line it's works, why could that be? (this issue is solved)
Any help would be great.

If you use happy with -i it will generate an info file. The file lists all the states that your parser has. It will also list all the possible transitions for each state. You can use this information to determine if the shift/reduce conflict is one you care about.
Information about invoking happy and conflicts:
http://www.haskell.org/happy/doc/html/sec-invoking.html
http://www.haskell.org/happy/doc/html/sec-conflict-tips.html
Below is some of the output of -i. I've removed all but State 17. You'll want to get a copy of this file so that you can properly debug the problem. What you see here is just to help talk about it:
-----------------------------------------------------------------------------
Info file generated by Happy Version 1.18.10 from FNC.y
-----------------------------------------------------------------------------
state 17 contains 4 shift/reduce conflicts.
-----------------------------------------------------------------------------
Grammar
-----------------------------------------------------------------------------
%start_parse -> Prop (0)
Prop -> Sentence '.' (1)
Sentence -> AtomSent (2)
Sentence -> CompSent (3)
AtomSent -> var (4)
CompSent -> '(' Sentence ')' (5)
CompSent -> Sentence Connective Sentence (6)
CompSent -> '¬' Sentence (7)
Connective -> and (8)
Connective -> or (9)
Connective -> "=>" (10)
Connective -> "<=>" (11)
-----------------------------------------------------------------------------
Terminals
-----------------------------------------------------------------------------
var { TokenVar $$ }
or { TokenOr }
and { TokenAnd }
'¬' { TokenNot }
"=>" { TokenImp }
"<=>" { TokenDImp }
'(' { TokenOB }
')' { TokenCB }
'.' { TokenEnd }
-----------------------------------------------------------------------------
Non-terminals
-----------------------------------------------------------------------------
%start_parse rule 0
Prop rule 1
Sentence rules 2, 3
AtomSent rule 4
CompSent rules 5, 6, 7
Connective rules 8, 9, 10, 11
-----------------------------------------------------------------------------
States
-----------------------------------------------------------------------------
State 17
CompSent -> Sentence . Connective Sentence (rule 6)
CompSent -> Sentence Connective Sentence . (rule 6)
or shift, and enter state 12
(reduce using rule 6)
and shift, and enter state 13
(reduce using rule 6)
"=>" shift, and enter state 14
(reduce using rule 6)
"<=>" shift, and enter state 15
(reduce using rule 6)
')' reduce using rule 6
'.' reduce using rule 6
Connective goto state 11
-----------------------------------------------------------------------------
Grammar Totals
-----------------------------------------------------------------------------
Number of rules: 12
Number of terminals: 9
Number of non-terminals: 6
Number of states: 19
That output basically says that it runs into a bit of ambiguity when it's looking at connectives. It turns out, the slides you linked mention this (Slide 11), "ambiguities are resolved through precedence ¬∧∨⇒⇔ or parentheses".
At this point, I would recommend looking at the shift/reduce conflicts and your desired precedences to see if the parser you have will do the right thing. If so, then you can safely ignore the warnings. If not, you have more work for yourself.

I can answer No. 2:
| c== '¬' == TokenNot : lexer cs --problem here
-- ^^
You have a == there where you should have a =.

Related

Parsing Decaf grammar in Antlr4

I am creating parser and lexer rules for Decaf programming language written in ANTLR4. There is a parser test file I am trying to run to get the parser tree for it by printing the visited nodes on the terminal window and paste them into D3_parser_tree.html class. The current parser tree is missing the right square brackets with the number 10 according to this testing file : class program { int i [10]; }
The error I am getting : mismatched input '10' expecting INT_LITERAL
I am not sure why I am getting this error although I have declared a lexer rule for INT_LITERAL and then called it in a parser rule within field_decl according to the given Decaf spec :
** Parser rules **
<program> → class Program ‘{‘ <field_decl>* <method_decl>* ‘}’
<field_decl> → <type> { <id> | <id> ‘[‘ <int_literal> ‘]’ }+, ;
<method_decl> → { <type> | void } <id> ( [ { <type> <id> }+, ] ) <block>
<digit> → 0 | 1 | 2 | … | 9
<block> → ‘{‘ <var_decl>* <statement>* ‘}’
<literal> → <int_literal> | <char_literal> | <bool_literal>
<hex_digit> → <digit> | a | b | c | … | f | A | B | C | … | F
<int_literal> → <decimal_literal> | <hex_literal>
<decimal_literal> → <digit> <digit>*
<hex_literal> → 0x <hex_digit> <hex_digit>*
Related Lexer rules :
NUMBER : [0-9]+;
fragment ALPHA : [_a-zA-Z0-9];
fragment DIGIT : [0-9];
fragment DECIMAL_LITERAL : DIGIT+;
CHAR_LITERAL : '\'' CHAR '\'';
STRING_LITERAL : '"' CHAR+ '"' ;
COMMENT : '//' ~('\n')* '\n' -> skip;
WS : (' ' | '\n' | '\t' | '\r') + -> skip;
Related Parser rules :
program : CLASS VAR LCURLYBRACE field_decl*method_decl* RCURLYBRACE EOF;
field_decl : data_type field ( COMMA field )* SEMICOLON;
Please let me know if you need further details & I appreciate your help a lot.
The following rules conflict:
VAR : ALPHA+;
...
NUMBER : [0-9]+;
...
INT_LITERAL : DECIMAL_LITERAL | HEX_LITERAL;
They all match 10, but the lexer will always choose VAR since that is the rule defined first.
This is just how ANTLR's lexer works: it tries to match the most characters as possible, and when two (or more) rules all match the same amount of characters, the one defined first "wins".
You will see that it parses correctly if you change field into:
field : VAR | VAR LSQUAREBRACE VAR RSQUAREBRACE;

How do I preform 'lookahead' in an OCaml lexer / how do I rollback a lexeme?

Well, I'm writing my first parser, in OCaml, and I immediately somehow managed to make one with an infinite-loop.
Of particular note, I'm trying to lex identifiers according to the rules of the Scheme specification (I have no idea what I'm doing, obviously) — and there's some language in there about identifiers requiring that they are followed by a delimiter. My approach, right now, is to have a delimited_identifier regex that includes one of the delimiter characters, that should not be consumed by the main lexer … and then once that's been matched, the reading of that lexeme is reverted by Sedlexing.rollback (well, my wrapper thereof), before being passed to a sublexer that only eats the actual identifier, hopefully leaving the delimiter in the buffer to be eaten as a different lexeme by the parent lexer.
I'm using Menhir and Sedlex, mostly synthesizing the examples from #smolkaj's ocaml-parsing example-repo and RWO's parsing chapter; here's the simplest reduction of my current parser and lexer:
%token LPAR RPAR LVEC APOS TICK COMMA COMMA_AT DQUO SEMI EOF
%token <string> IDENTIFIER
(* %token <bool> BOOL *)
(* %token <int> NUM10 *)
(* %token <string> STREL *)
%start <Parser.AST.t> program
%%
program:
| p = list(expression); EOF { p }
;
expression:
| i = IDENTIFIER { Parser.AST.Atom i }
%%
… and …
(** Regular expressions *)
let newline = [%sedlex.regexp? '\r' | '\n' | "\r\n" ]
let whitespace = [%sedlex.regexp? ' ' | newline ]
let delimiter = [%sedlex.regexp? eof | whitespace | '(' | ')' | '"' | ';' ]
let digit = [%sedlex.regexp? '0'..'9']
let letter = [%sedlex.regexp? 'A'..'Z' | 'a'..'z']
let special_initial = [%sedlex.regexp?
'!' | '$' | '%' | '&' | '*' | '/' | ':' | '<' | '=' | '>' | '?' | '^' | '_' | '~' ]
let initial = [%sedlex.regexp? letter | special_initial ]
let special_subsequent = [%sedlex.regexp? '+' | '-' | '.' | '#' ]
let subsequent = [%sedlex.regexp? initial | digit | special_subsequent ]
let peculiar_identifier = [%sedlex.regexp? '+' | '-' | "..." ]
let identifier = [%sedlex.regexp? initial, Star subsequent | peculiar_identifier ]
let delimited_identifier = [%sedlex.regexp? identifier, delimiter ]
(** Swallow whitespace and comments. *)
let rec swallow_atmosphere buf =
match%sedlex buf with
| Plus whitespace -> swallow_atmosphere buf
| ";" -> swallow_comment buf
| _ -> ()
and swallow_comment buf =
match%sedlex buf with
| newline -> swallow_atmosphere buf
| any -> swallow_comment buf
| _ -> assert false
(** Return the next token. *)
let rec token buf =
swallow_atmosphere buf;
match%sedlex buf with
| eof -> EOF
| delimited_identifier ->
Sedlexing.rollback buf;
identifier buf
| '(' -> LPAR
| ')' -> RPAR
| _ -> illegal buf (Char.chr (next buf))
and identifier buf =
match%sedlex buf with
| _ -> IDENTIFIER (Sedlexing.Utf8.lexeme buf)
(Yes, it's basically a no-op / the simplest thing possible rn. I'm trying to learn! :x)
Unfortunately, this combination results in an infinite loop in the parsing automaton:
State 0:
Lookahead token is now IDENTIFIER (1-1)
Shifting (IDENTIFIER) to state 1
State 1:
Lookahead token is now IDENTIFIER (1-1)
Reducing production expression -> IDENTIFIER
State 5:
Shifting (IDENTIFIER) to state 1
State 1:
Lookahead token is now IDENTIFIER (1-1)
Reducing production expression -> IDENTIFIER
State 5:
Shifting (IDENTIFIER) to state 1
State 1:
...
I'm new to parsing and lexing and all this; any advice would be welcome. I'm sure it's just a newbie mistake, but …
Thanks!
As said before, implementing too much logic inside the lexer is a bad idea.
However, the infinite loop does not come from the rollback but from your definition of identifier:
identifier buf =
match%sedlex buf with
| _ -> IDENTIFIER (Sedlexing.Utf8.lexeme buf)
within this definition _ matches the shortest possible words in the language consisting of all possible characters. In other words, _ always matches the empty word μ without consuming any part of its input, sending the parser into an infinite loop.

How does a parser solves shift/reduce conflict?

I have a grammar for arithmetic expression which solves number of expression (one per line) in a text file. While compiling YACC I am getting message 2 shift reduce conflicts. But my calculations are proper. If parser is giving proper output how does it resolves the shift/reduce conflict. And In my case is there any way to solve it in YACC Grammar.
YACC GRAMMAR
Calc : Expr {printf(" = %d\n",$1);}
| Calc Expr {printf(" = %d\n",$2);}
| error {yyerror("\nBad Expression\n ");}
;
Expr : Term { $$ = $1; }
| Expr '+' Term { $$ = $1 + $3; }
| Expr '-' Term { $$ = $1 - $3; }
;
Term : Fact { $$ = $1; }
| Term '*' Fact { $$ = $1 * $3; }
| Term '/' Fact { if($3==0){
yyerror("Divide by Zero Encountered.");
break;}
else
$$ = $1 / $3;
}
;
Fact : Prim { $$ = $1; }
| '-' Prim { $$ = -$2; }
;
Prim : '(' Expr ')' { $$ = $2; }
| Id { $$ = $1; }
;
Id :NUM { $$ = yylval; }
;
What change should I do to remove such conflicts in my grammar ?
Bison/yacc resolves shift-reduce conflicts by choosing to shift. This is explained in the bison manual in the section on Shift-Reduce conflicts.
Your problem is that your input is just a series of Exprs, run together without any delimiter between them. That means that:
4 - 2
could be one expression (4-2) or it could be two expressions (4, -2). Since bison-generated parsers always prefer to shift, the parser will choose to parse it as one expression, even if it were typed on two lines:
4
-2
If you want to allow users to type their expressions like that, without any separator, then you could either live with the conflict (since it is relatively benign) or you could codify it into your grammar, but that's quite a bit more work. To put it into the grammar, you need to define two different types of Expr: one (which is the one you use at the top level) cannot start with an unary minus, and the other one (which you can use anywhere else) is allowed to start with a unary minus.
I suspect that what you really want to do is use newlines or some other kind of expression separator. That's as simple as passing the newline through to your parser and changing Calc to Calc: | Calc '\n' | Calc Expr '\n'.
I'm sure that this appears somewhere else on SO, but I can't find it. So here is how you disallow the use of unary minus at the beginning of an expression, so that you can run expressions together without delimiters. The non-terminals starting n_ cannot start with a unary minus:
input: %empty | input n_expr { /* print $2 */ }
expr: term | expr '+' term | expr '-' term
n_expr: n_term | n_expr '+' term | n_expr '-' term
term: factor | term '*' factor | term '/' factor
n_term: value | n_term '+' factor | n_term '/' factor
factor: value | '-' factor
value: NUM | '(' expr ')'
That parses the same language as your grammar, but without generating the shift-reduce conflict. Since it parses the same language, the input
4
-2
will still be parsed as a single expression; to get the expected result you would need to type
4
(-2)

Managing position information with Alex and Happy

I'm learning to use Alex and Happy to write a small compiler. I want to maintain line and column information for my AST nodes so that I can provide meaningful error messages to the user. To illustrate how I plan to do it, I wrote a small example (see code below), and I'd like to know if the way I approached the problem (having AlexPosn attached to the tokens, attaching a polymorphic attribute field to AST nodes, using tkPos and astAttr) is good style or if there are better ways to handle position information.
Lexer.x:
{
module Lexer where
}
%wrapper "posn"
$white = [\ \t\n]
tokens :-
$white+ ;
[xX] { \pos s -> MkToken pos X }
"+" { \pos s -> MkToken pos Plus }
"*" { \pos s -> MkToken pos Times }
"(" { \pos s -> MkToken pos LParen }
")" { \pos s -> MkToken pos RParen }
{
data Token = MkToken AlexPosn TokenClass
deriving (Show, Eq)
data TokenClass = X
| Plus
| Times
| LParen
| RParen
deriving (Show, Eq)
tkPos :: Token -> (Int, Int)
tkPos (MkToken (AlexPn _ line col) _) = (line, col)
}
Parser.y:
{
module Parser where
import Lexer
}
%name simple
%tokentype { Token }
%token
'(' { MkToken _ LParen }
')' { MkToken _ RParen }
'+' { MkToken _ Plus }
'*' { MkToken _ Times }
x { MkToken _ X }
%%
Expr : Term '+' Expr { NAdd $1 $3 (astAttr $1) }
| Term { $1 }
Term : Factor '*' Term { NMul $1 $3 (astAttr $1) }
| Factor { $1 }
Factor : x { NX (tkPos $1) }
| '(' Expr ')' { $2 }
{
data AST a = NX a
| NMul (AST a) (AST a) a
| NAdd (AST a) (AST a) a
deriving (Show, Eq)
astAttr :: AST a -> a
astAttr (NX a) = a
astAttr (NMul _ _ a) = a
astAttr (NAdd _ _ a) = a
happyError :: [Token] -> a
happyError _ = error "parse error"
}
Main.hs:
module Main where
import Lexer
import Parser
main :: IO ()
main = do
s <- getContents
let toks = alexScanTokens s
print $ simple toks
I personally would be pretty ok with the style you've described. However, it is very manual and I was hoping to at least provide one alternative that might be easier to manage.
If you look a little further down the documentation for alex wrappers, you'll notice that the monad and monadstate wrappers both contain position information. The downside is that you now have the entire thing wrapped in a monad and it complicates the parser slightly. However, by wrapping it in a monad, the result of the parse is an Alex a which means you have full access to the line and column information when you create your ast nodes. Now this simply removes some of the boiler plate from the lexer and doesn't do much more.
By doing this, you could also carry around the AlexState with your token, but that might be unnecessary.
If you need help actually fixing the parser to handle a monad/monadstate wrapper, I wrote a response on how I managed to get it working here: How to use an Alex monadic lexer with Happy?

Why does ANTLR not parse the entire input?

I am quite new to ANTLR, so this is likely a simple question.
I have defined a simple grammar which is supposed to include arithmetic expressions with numbers and identifiers (strings that start with a letter and continue with one or more letters or numbers.)
The grammar looks as follows:
grammar while;
#lexer::header {
package ConFreeG;
}
#header {
package ConFreeG;
import ConFreeG.IR.*;
}
#parser::members {
}
arith:
term
| '(' arith ( '-' | '+' | '*' ) arith ')'
;
term returns [AExpr a]:
NUM
{
int n = Integer.parseInt($NUM.text);
a = new Num(n);
}
| IDENT
{
a = new Var($IDENT.text);
}
;
fragment LOWER : ('a'..'z');
fragment UPPER : ('A'..'Z');
fragment NONNULL : ('1'..'9');
fragment NUMBER : ('0' | NONNULL);
IDENT : ( LOWER | UPPER ) ( LOWER | UPPER | NUMBER )*;
NUM : '0' | NONNULL NUMBER*;
fragment NEWLINE:'\r'? '\n';
WHITESPACE : ( ' ' | '\t' | NEWLINE )+ { $channel=HIDDEN; };
I am using ANTLR v3 with the ANTLR IDE Eclipse plugin. When I parse the expression (8 + a45) using the interpreter, only part of the parse tree is generated:
Why does the second term (a45) not get parsed? The same happens if both terms are numbers.
You'll want to create a parser rule that has an EOF (end of file) token in it so that the parser will be forced to go through the entire token stream.
Add this rule to your grammar:
parse
: arith EOF
;
and let the interpreter start at that rule instead of the arith rule:

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