How to resolve the no viable input error in racket? - parsing

I've been trying to execute the following program in a parser
#lang racket
( define ( highest-number xs )
( define ( max x1 x2 )
( if ( > x1 x2) x1 x2 ) )
( foldl max ( first xs ) ( rest xs ) )
The error that is generated is as follows :
line 3:4 no viable alternative at input '( define'
These are my rules :
grammar hello;
program
: defOrExpr+ EOF
;
defOrExpr
: definition
| expr
| testCase
| libraryRequire
;
definition
: '(' 'define' '(' name NAME+ ')' expr ')'
| '(' 'define' name expr ')'
| '(' 'define-struct' name '(' name* ')' ')'
;

(As Bart mentioned, it's much easier to help if there's a buildable grammar and sample input to reproduce your problem.)
In this case, I think the problem is fairly obvious.
Your definition rule does not contain an alternative that allows for a nested definition, and your sample input has a nested definition.
I'm not familiar enough with Racket to suggest an alternative to address the issue, but that's why you're getting your error. (it's on the ( define ( max x1 x2 ) part, not the first define)

Related

EBNF with comma repetition

I find myself doing the following quite frequently as allowing one multiple entries separated by a comma:
( function | expression ) ( ',' ( function | expression ))*
Is there a more compact way to do this? Ideally I'd just like to be able to do something along the lines of:
( function | expression ) [,...]
Or:
( function | expression ',')*
By the way, I am using this as a validator: https://www.bottlecaps.de/rr/ui#_Production
The whole grammar I am trying to 'clean up' is the following:
AGGREGATION
::= 'GROUP BY' ( GROUPING_ROWS | PIVOT )?
PIVOT
::= 'PIVOT(' AXIS_EXPR (AXIS_EXPR ',' )? ')'
AXIS_EXPR
::= expr ( 'AS'? alias )? 'ON' ( 'ROWS' | 'COLS' ) ( 'HAVING' expr )? ( 'ORDER BY' expr ( 'ASC' | 'DESC' )? )? ( 'LIMIT' num 'PERCENT'? )?
GROUPING_ROWS
::= 'GROUPING_ROWS(' GROUPING_EXPR (GROUPING_EXPR ',' )? ')'
GROUPING_EXPR
::= NAME_OR_POS 'SUBTOTAL' 'S'? GROUPING_EXPR_SUBTOTAL (',' GROUPING_EXPR_SUBTOTAL)*
GROUPING_EXPR_SUBTOTAL
::= NAME_OR_POS ':' AGGREGATED_CALCULATION ( ',' AGGREGATED_CALCULATION )*
NAME_OR_POS
::= ( name | pos )
AGGREGATED_CALCULATION
::= ( aggregation_function | aggregation_expression ) ( 'AS'? alias)?
And as an example of the construct I find myself using all the time:
( function | expression ) ( ',' ( function | expression ))*
Is there a more compact way to do this?
Other than introducing "helper rules" like this:
rule
: atom_list
;
atom_list
: atom (',' atom)*
;
atom
: function
| expression
;
the answer is: no, there is no shorter way to write a (',' a)* into something like (a ',')* with ANTLR.
If you're repeating function | expression a lot, at the very least make a separate rule of those alternatives.

Is there a smart way to use different types of brackets with ANTLR

I'm making a Lisp parser and I wanted to accept the following line:
(defvar a [+ 2 1] )
But not this one:
(defvar a [+ 2 1) ]
The solution i came up with was something like this:
define_var: ('(' KW_DEFVAR ID expression ')' ) | ('[' KW_DEFVAR ID expression ']');
It works, but for longer expressions this is not viable.
You could move the tokens/rules in common in a sub rule like this:
define_var
: '(' define_var_subrule ')'
| '[' define_var_subrule ']'
;
define_var_subrule
: KW_DEFVAR ID expression
;

Antlr4: Another "No Viable Alternative Error"

I have checked similar questions surrounding this issue but none seems to provide a solution to my version of the problem.
I just started Antlr4 recently and all has been going nicely until I hit this particular roadblock.
My grammar is a basic math expression grammar but for some reason I noticed the generated parser(?) is unable to walk from paser-rule "equal" to paser-rule "expr", in order to reach lexer-rule "NAME".
grammar MathCraze;
NUM : [0-9]+ ('.' [0-9]+)?;
WS : [ \t]+ -> skip;
NL : '\r'? '\n' -> skip;
NAME: [a-zA-Z_][a-zA-Z_0-9]*;
ADD: '+';
SUB : '-';
MUL : '*';
DIV : '/';
POW : '^';
equal
: add # add1
| NAME '=' equal # assign
;
add
: mul # mul1
| add op=('+'|'-') mul # addSub
;
mul
: exponent # power1
| mul op=('*'|'/') exponent # mulDiv
;
exponent
: expr # expr1
| expr '^' exponent # power
;
expr
: NUM # num
| NAME # name
| '(' add ')' # parens
;
If I pass a word as input, sth like "variable", the parser throws the error above, but if I pass a number as input (say "78"), the parser walks the tree successfully (i.e, from rule "equal" to "expr").
equal equal
| |
add add
| |
mul mul
| |
exponent exponent
| |
expr expr
| |
NUM NAME
| |
"78" # No Error "variable" # Error! Tree walk doesn't reach here.
I've checked for every type of ambiguity I know of, so I'm probably missing something here.
I'm using Antlr5.6 by the way and I will appreciate if this problem gets solved. Thanks in advance.
Your style of expression hierarchy is the one we use in parsers written by hand or in ANTLR v3, from low to high precedence.
As Raven said, ANTLR 4 is much more powerful. Note the <assoc = right> specification in the power rule, which is usually right-associative.
grammar Question;
question
: line+ EOF
;
line
: expr NL
| assign NL
;
assign
: NAME '=' expr # assignSingle
| NAME '=' assign # assignMulti
;
expr // from high to low precedence
: <assoc = right> expr '^' expr # power
| expr op=( '*' | '/' ) expr # mulDiv
| expr op=( '+' | '-' ) expr # addSub
| '(' expr ')' # parens
| atom_r # atom
;
atom_r
: NUM
| NAME
;
NAME: [a-zA-Z_][a-zA-Z_0-9]*;
NUM : [0-9]+ ('.' [0-9]+)?;
WS : [ \t]+ -> skip;
NL : [\r\n]+ ;
Run with the -gui option to see the parse tree :
$ echo $CLASSPATH
.:/usr/local/lib/antlr-4.6-complete.jar
$ alias grun
alias grun='java org.antlr.v4.gui.TestRig'
$ grun Question question -gui data.txt
and this data.txt file :
variable
78
a + b * c
a * b + c
a = 8 + (6 * 9)
a ^ b
a ^ b ^ c
7 * 2 ^ 5
a = b = c = 88
.
Added
Using your original grammar and starting with the equal rule, I have the following error :
$ grun Q2 equal -tokens data.txt
[#0,0:7='variable',<NAME>,1:0]
[#1,9:10='78',<NUM>,2:0]
...
[#41,89:88='<EOF>',<EOF>,10:0]
line 2:0 no viable alternative at input 'variable78'
If I start with rule expr, there is no error :
$ grun Q2 expr -tokens data.txt
[#0,0:7='variable',<NAME>,1:0]
...
[#41,89:88='<EOF>',<EOF>,10:0]
$
Run grun with the -gui option and you'll see the difference :
running with expr, the input token variable is catched in NAME, rule expr is satisfied and terminates;
running with equal it's all in error. The parser tries the first alternative equal -> add -> mul -> exponent -> expr -> NAME => OK. It consumes the token variable and tries to do something with the next token 78. It rolls back in each rule, see if it can do something with the alt of rule, but each alt requires an operator. Thus it arrives in equal and starts again with the token variable, this time using the alt | NAME '='. NAME consumes the token, then the rule requires '=', but the input is 78 and does not satisfies it. As there is no other choice, it says there is no viable alternative.
$ grun Q2 equal -tokens data.txt
[#0,0:7='variable',<NAME>,1:0]
[#1,8:7='<EOF>',<EOF>,1:8]
line 1:8 no viable alternative at input 'variable'
If variable is the only token, same reasoning : first alternative equal -> add -> mul -> exponent -> expr -> NAME => OK, consumes variable, back to equal, tries the alt which requires '=', but the input is at EOF. That's why it says there is no viable alternative.
$ grun Q2 equal -tokens data.txt
[#0,0:1='78',<NUM>,1:0]
[#1,2:1='<EOF>',<EOF>,1:2]
If 78 is the only token, do the same reasoning : first alternative equal -> add -> mul -> exponent -> expr -> NUM => OK, consumes 78, back to equal. The alternative is not an option. Satisfied ? oops, what about EOF.
Now let's add a NUM alt to equal :
equal
: add # add1
| NAME '=' equal # assign
| NUM '=' equal # assignNum
;
$ grun Q2 equal -tokens data.txt
[#0,0:1='78',<NUM>,1:0]
[#1,2:1='<EOF>',<EOF>,1:2]
line 1:2 no viable alternative at input '78'
First alternative equal -> add -> mul -> exponent -> expr -> NUM => OK, consumes 78, back to equal. Now there is also an alt for NUM, starts again, this time using the alt | NUM '='. NUM consumes the token 78,
then the parser requires '=', but the input is at EOF, hence the message.
Now let's add a new rule with EOF and let's run the grammar from all :
all : equal EOF ;
$ grun Q2 all -tokens data.txt
[#0,0:1='78',<NUM>,1:0]
[#1,2:1='<EOF>',<EOF>,1:2]
$ grun Q2 all -tokens data.txt
[#0,0:7='variable',<NAME>,1:0]
[#1,8:7='<EOF>',<EOF>,1:8]
The input corresponds to the grammar, and there is no more message.
Although I can't answer your question about why the parser can't reach NAME in expr I'd like to point out that with Antlr4 you can use direct left recursion in your rule specification which makes your grammar more compact and omproves readability.
With that in mind your grammar could be rewritten as
math:
assignment
| expression
;
assignment:
ID '=' (assignment | expression)
;
expression:
expression '^' expression
| expression ('*' | '/') expression
| expression ('+' | '-') expression
| NAME
| NUM
;
That grammar hapily takes a NAME as part of an expression so I guess it would solve your problem.
If you're really interested in why it didn't work with your grammar then I'd first check if the lexer has matched the input into the expected tokens. Afterwards I would have a look at the parse tree to see what the parser is making of the given token sequence and then trying to do the parsing manually accoding to your grammar and during that you should be able to find the point at which the parser does something different from what you'd expect it to do.

yacc shift/reduce conflict when parse C ++ template arguments

I have a YACC grammar for parsing expressions in C++. Here is lite version:
// yacc.y
%token IDENT
%%
expr:
call_expr
| expr '<' call_expr
| expr '>' call_expr
;
call_expr:
IDENT
| '(' expr ')'
| IDENT '<' args '>' '(' args ')'
;
args:
IDENT
| args ',' IDENT
;
%%
When I want to support function call with template arguments, I got a shift/reduce conflict.
When we got input IDENT '<' IDENT, yacc doesn't know whether we should shift or reduce.
I want IDENT '<' args '>' '(' args ')' got higher precedence level than expr '<' call_expr, so I can parse the following exprs.
x < y
f<x>(a,b)
f<x,y>(a,b) < g<x,y>(c,d)
I see C++/C# both support this syntax. Is there any way to solve this problem with yacc?
How do I modify the .y file?
Thank you!
You want the -v option to yacc/bison. It will give you a .output file with all the info about the generated shift/reduce parser. With your grammar, bison gives you:
State 1 conflicts: 1 shift/reduce
:
state 1
4 call_expr: IDENT .
6 | IDENT . '<' args '>' '(' args ')'
'<' shift, and go to state 5
'<' [reduce using rule 4 (call_expr)]
$default reduce using rule 4 (call_expr)
which shows you where the problem is. After seeing an IDENT, when the next token is <, it doesn't know if it should reduce that call_expr (to ultimately match the rule expr: expr '<' call_expr) or if it should shift to match rule 6.
Parsing this with only 1 token lookahead is hard, as you have two distinct meanings of the token < (less-than or open angle bracket), and which is meant depends on the later tokens.
This case is actually even worse, as it is ambiguous, as an input like
a < b > ( c )
is probably a template call with both args lists being singletons, but it could also be
( a < b ) > ( c )
so just unfactoring the grammar won't help. Your best bet is to use a more powerful parsing method, like bison's %glr-parser option or btyacc

Practical solution to fix a Grammar Problem

We have little snippets of vb6 code (the only use a subset of features) that gets wirtten by non-programmers. These are called rules. For the people writing these they are hard to debug so somebody wrote a kind of add hoc parser to be able to evaluete the subexpressions and thereby show better where the problem is.
This addhoc parser is very bad and does not really work woll. So Im trying to write a real parser (because im writting it by hand (no parser generator I could understand with vb6 backends) I want to go with recursive decent parser). I had to reverse-engineer the grammer because I could find anything. (Eventully I found something http://www.notebar.com/GoldParserEngine.html but its LALR and its way bigger then i need)
Here is the grammer for the subset of VB.
<Rule> ::= expr rule | e
<Expr> ::= ( expr )
| Not_List CompareExpr <and_or> expr
| Not_List CompareExpr
<and_or> ::= Or | And
<Not_List> ::= Not Not_List | e
<CompareExpr> ::= ConcatExpr comp CompareExpr
|ConcatExpr
<ConcatExpr> ::= term term_tail & ConcatExpr
|term term_tail
<term> ::= factor factor_tail
<term_tail> ::= add_op term term_tail | e
<factor> ::= add_op Value | Value
<factor_tail> ::= multi_op factor factor_tail | e
<Value> ::= ConstExpr | function | expr
<ConstExpr> ::= <bool> | number | string | Nothing
<bool> ::= True | False
<Nothing> ::= Nothing | Null | Empty
<function> ::= id | id ( ) | id ( arg_list )
<arg_list> ::= expr , arg_list | expr
<add_op> ::= + | -
<multi_op> ::= * | /
<comp> ::= > | < | <= | => | =< | >= | = | <>
All in all it works pretty good here are some simple examples:
my_function(1, 2 , 3)
looks like
(Programm
(rule
(expr
(Not_List)
(CompareExpr
(ConcatExpr
(term
(factor
(value
(function
my_function
(arg_list
(expr
(Not_List)
(CompareExpr
(ConcatExpr (term (factor (value 1))) (term_tail))))
(arg_list
(expr
(Not_List)
(CompareExpr
(ConcatExpr (term (factor (value 2))) (term_tail))))
(arg_list
(expr
(Not_List)
(CompareExpr
(ConcatExpr (term (factor (value 3))) (term_tail))))
(arg_list))))))))
(term_tail))))
(rule)))
Now whats my problem?
if you have code that looks like this (( true OR false ) AND true) I have a infinit recursion but the real problem is that in the (true OR false) AND true (after the first ( expr ) ) is understood as only (true or false).
Here is the Parstree:
So how to solve this. Should I change the grammer somehow or use some implmentation hack?
Something hard exmplale in case you need it.
(( f1 OR f1 ) AND (( f3="ALL" OR f4="test" OR f5="ALL" OR f6="make" OR f9(1, 2) ) AND ( f7>1 OR f8>1 )) OR f8 <> "")
You have several issues that I see.
You are treating OR and AND as equal precedence operators. You should have separate rules for OR, and for AND. Otherwise you will the wrong precedence (therefore evaluation) for the expression A OR B AND C.
So as a first step, I'd revise your rules as follows:
<Expr> ::= ( expr )
| Not_List AndExpr Or Expr
| Not_List AndExpr
<AndExpr> ::=
| CompareExpr And AndExpr
| Not_List CompareExpr
Next problem is that you have ( expr ) at the top level of your list. What if I write:
A AND (B OR C)
To fix this, change these two rules:
<Expr> ::= Not_List AndExpr Or Expr
| Not_List AndExpr
<Value> ::= ConstExpr | function | ( expr )
I think your implementation of Not is not appropriate. Not is an operator,
just with one operand, so its "tree" should have a Not node and a child which
is the expression be Notted. What you have a list of Nots with no operands.
Try this instead:
<Expr> ::= AndExpr Or Expr
| AndExpr
<Value> ::= ConstExpr | function | ( expr ) | Not Value
I haven't looked, but I think VB6 expressions have other messy things in them.
If you notice, the style of Expr and AndExpr I have written use right recursion to avoid left recursion. You should change your Concat, Sum, and Factor rules to follow a similar style; what you have is pretty complicated and hard to follow.
If they are just creating snippets then perhaps VB5 is "good enough" for creating them. And if VB5 is good enough, the free VB5 Control Creation Edition might be worth tracking down for them to use:
http://www.thevbzone.com/vbcce.htm
You could have them start from a "test harness" project they add snippets to, and they can even test them out.
With a little orientation this will probably prove much more practical than hand crafting a syntax analyzer, and a lot more useful since they can test for more than correct syntax.
Where VB5 is lacking you might include a static module in the "test harness" that provides a rough and ready equivalent of Split(), Replace(), etc:
http://support.microsoft.com/kb/188007

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