Left-recursive grammars in operator precedence parsing2019 Community Moderator ElectionRecursive grammars in FParsecParsing method arguments with FParsecHow to handle left recursion in FParsecParsing numbers in FParsecParse case-insensitive operators using OperatorPrecedenceParserParsing “x y z” with the precedence of multiplyParsing int or float with FParsecHow to insert an indentation check into the operator precedence parser?Confusion as to how to integrate OperatorPrecedenceParser with mineParsing in to a recursive data structure
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Left-recursive grammars in operator precedence parsing
2019 Community Moderator ElectionRecursive grammars in FParsecParsing method arguments with FParsecHow to handle left recursion in FParsecParsing numbers in FParsecParse case-insensitive operators using OperatorPrecedenceParserParsing “x y z” with the precedence of multiplyParsing int or float with FParsecHow to insert an indentation check into the operator precedence parser?Confusion as to how to integrate OperatorPrecedenceParser with mineParsing in to a recursive data structure
I have a left recursive grammar. My AST looks something like:
...
and Expr = BinaryExpr of BinaryExpr
and BinaryExpr = Expr * BinaryOperator * Expr
and BinaryOperator = Plus
...
I was planning on using the operator precedence parser for this, for example:
let exprOpp = new OperatorPrecedenceParser<Expr,unit,unit>()
let pBinaryExpr = exprOpp.ExpressionParser
exprOpp.TermParser <- pExpr
let consBinExpr op x y = (x, op, y) |> BinaryExpr
exprOpp.AddOperator(InfixOperator("+", ws, 1, Associativity.Left, (consBinExpr Plus)))
where pExpr is a parser for Expr.
The left recursion causes a stack overflow. I was wondering whether there was a particular FParsec-way of dealing with this while still using the OperatorPrecedenceParser?
Thanks!
EDIT:
The issue may be coming from the fact that pExpr forwards all calls to another parser pExprRef.
let pExpr, pExprRef = createParserForwardedToRef()
and after the operator precedence parser, the reference is defined:
do pExprRef :=
choice [pBinaryExpr, <other-parsers...>];
f# fparsec
asked yesterday
falkmarfalkmar
11
add a comment |
I have a left recursive grammar. My AST looks something like:
...
and Expr = BinaryExpr of BinaryExpr
and BinaryExpr = Expr * BinaryOperator * Expr
and BinaryOperator = Plus
...
I was planning on using the operator precedence parser for this, for example:
let exprOpp = new OperatorPrecedenceParser<Expr,unit,unit>()
let pBinaryExpr = exprOpp.ExpressionParser
exprOpp.TermParser <- pExpr
let consBinExpr op x y = (x, op, y) |> BinaryExpr
exprOpp.AddOperator(InfixOperator("+", ws, 1, Associativity.Left, (consBinExpr Plus)))
where pExpr is a parser for Expr.
The left recursion causes a stack overflow. I was wondering whether there was a particular FParsec-way of dealing with this while still using the OperatorPrecedenceParser?
Thanks!
EDIT:
The issue may be coming from the fact that pExpr forwards all calls to another parser pExprRef.
let pExpr, pExprRef = createParserForwardedToRef()
and after the operator precedence parser, the reference is defined:
do pExprRef :=
choice [pBinaryExpr, <other-parsers...>];
f# fparsec
asked yesterday
falkmarfalkmar
11
Your example would never terminate; you'd need another option inExprthat represents a single value. As it is, your example types are infinitely recursive.
– rmunn
yesterday
add a comment |
I have a left recursive grammar. My AST looks something like:
...
and Expr = BinaryExpr of BinaryExpr
and BinaryExpr = Expr * BinaryOperator * Expr
and BinaryOperator = Plus
...
I was planning on using the operator precedence parser for this, for example:
let exprOpp = new OperatorPrecedenceParser<Expr,unit,unit>()
let pBinaryExpr = exprOpp.ExpressionParser
exprOpp.TermParser <- pExpr
let consBinExpr op x y = (x, op, y) |> BinaryExpr
exprOpp.AddOperator(InfixOperator("+", ws, 1, Associativity.Left, (consBinExpr Plus)))
where pExpr is a parser for Expr.
The left recursion causes a stack overflow. I was wondering whether there was a particular FParsec-way of dealing with this while still using the OperatorPrecedenceParser?
Thanks!
EDIT:
The issue may be coming from the fact that pExpr forwards all calls to another parser pExprRef.
let pExpr, pExprRef = createParserForwardedToRef()
and after the operator precedence parser, the reference is defined:
do pExprRef :=
choice [pBinaryExpr, <other-parsers...>];
f# fparsec
asked yesterday
falkmarfalkmar
11
I have a left recursive grammar. My AST looks something like:
...
and Expr = BinaryExpr of BinaryExpr
and BinaryExpr = Expr * BinaryOperator * Expr
and BinaryOperator = Plus
...
I was planning on using the operator precedence parser for this, for example:
let exprOpp = new OperatorPrecedenceParser<Expr,unit,unit>()
let pBinaryExpr = exprOpp.ExpressionParser
exprOpp.TermParser <- pExpr
let consBinExpr op x y = (x, op, y) |> BinaryExpr
exprOpp.AddOperator(InfixOperator("+", ws, 1, Associativity.Left, (consBinExpr Plus)))
where pExpr is a parser for Expr.
The left recursion causes a stack overflow. I was wondering whether there was a particular FParsec-way of dealing with this while still using the OperatorPrecedenceParser?
Thanks!
EDIT:
The issue may be coming from the fact that pExpr forwards all calls to another parser pExprRef.
let pExpr, pExprRef = createParserForwardedToRef()
and after the operator precedence parser, the reference is defined:
do pExprRef :=
choice [pBinaryExpr, <other-parsers...>];
f# fparsec
f# fparsec
asked yesterday
falkmarfalkmar
11
asked yesterday
falkmarfalkmar
11
edited yesterday
falkmar
asked yesterday
falkmarfalkmar
11
asked yesterday
falkmarfalkmar
11
asked yesterday
falkmarfalkmar
11
11
Your example would never terminate; you'd need another option inExprthat represents a single value. As it is, your example types are infinitely recursive.
– rmunn
yesterday
add a comment |
Your example would never terminate; you'd need another option inExprthat represents a single value. As it is, your example types are infinitely recursive.
– rmunn
yesterday
Your example would never terminate; you'd need another option in
Expr that represents a single value. As it is, your example types are infinitely recursive.– rmunn
yesterday
Your example would never terminate; you'd need another option in
Expr that represents a single value. As it is, your example types are infinitely recursive.– rmunn
yesterday
add a comment |
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Your example would never terminate; you'd need another option in
Exprthat represents a single value. As it is, your example types are infinitely recursive.– rmunn
yesterday