I am coding a parser of expression and visualization of it, which means every step of the recursive descent parsing or construction of AST will be visualized like a tiny version of VisuAlgo
// Expression grammer
Goal -> Expr
Expr -> Term + Term
| Expr - Term
| Term
Term -> Term * Factor
| Term / Factor
| Factor
Factor -> (Expr)
| num
| name
So I am wondering what data structure can be easily used for storing every step of constructing AST and how to implement the visualization of every step of constructing AST. Well, I have searched some similar questions and implement a recursive descent parser before, but just can't get a way to figure this out. I will be appreciated if anyone can help me.
This SO answer shows how to parse and build a tree as you parse
You could easily just the print the tree or tree fragments at each point where they are created.
To print the tree, just walk it recursively and print the nodes with indentation equal to depth of recursion.
Related
Currently I'm trying to build a parser for VHDL which
has some of the problems C++-Parsers have to face.
The context-free grammar of VHDL produces a parse
forest rather than a single parse tree because of it's
ambiguity regarding function calls and array subscriptions
foo := fun(3) + arr(5);
This assignment can only be parsed unambiguous if the parser
would carry around a hirachically, type-aware symbol table
which it'd use to resolve the ambiguities somewhat on-the-fly.
I don't want to do this, because for statements like the
aforementioned, the parse forest would not grow exponentially, but
rather linear depending on the amount of function calls and
array subscriptions.
(Except, of course, one would torture the parser with statements like)
foo := fun(fun(fun(fun(fun(4)))));
Since bison forces the user to just create one single parse-tree,
I used %merge attributes to collect all subtrees recursively and
added those subtrees under so called AMBIG nodes in the singleton
AST.
The result looks like this.
In order to produce the above, I parsed the token stream "I=I(N);".
The substance of the grammar I used inside the parse.y file, is
collected below. It tries to resemble the ambiguous parts of VHDL:
start: prog
;
/* I cut out every semantic action to make this
text more readable */
prog: assignment ';'
| prog assignment ';'
;
assignment: 'I' '=' expression
;
expression: function_call %merge <stmtmerge2>
| array_indexing %merge <stmtmerge2>
| 'N'
;
function_call: 'I' '(' expression ')'
| 'I'
;
array_indexing: function_call '(' expression ')' %merge <stmtmerge>
| 'I' '(' expression ')' %merge <stmtmerge>
;
The whole sourcecode can be read at this github repository.
And now, let's get down to the actual Problem.
As you can see in the generated parse tree above,
the nodes FCALL1 and ARRIDX1 refer to the same
single node EXPR1 which in turn refers to N1 twice.
This, by all means, should not have happened and I don't
know why. Instead there should be the paths
FCALL1 -> EXPR2 -> N2
ARRIDX1 -> EXPR1 -> N1
Do you have any idea why bison reuses the aforementioned
nodes?
I also wrote a bugreport on the official gnu mailing
list for bison, without a reply to this point though.
Unfortunately, due to the restictions for new stackoverflow
users, I can't provide no link to this bug report...
That behaviour is expected.
expression can be unambiguously reduced, and that reduced value is used by both possible ambiguous reductions which include the value. Remember that GLR, like LR, is a left-to-right parser. When a reduction action is executed, all of the child reductions have already happened. The effect is not different from the use of a terminal in a right-hand side; the terminal will not be artificially copied in order to produce different instances in the ambiguous productions which use it.
For most people, this would be a feature rather than a bug, and I don't mean that as a joke. Without the graph-structured stack, GLR has exponential run-time. If you really want to do a deep copy of shared AST nodes when you merge parse trees, you will have to do it yourself, but I suggest that you find a way to make use of the fact that the parse forest is really an directed acyclic graph rather than a tree; you will probably be able to take advantage of the lack of duplication.
I have been trying create my parser for expression with variables and simplify them to quadratic expression form.
This is my parser grammar:
Exercise : Expr '=' Expr
Expr : Term [+-] Expr | Term
Term : Factor [*/] Term | Factor
Factor: Atom [^] Factor | Atom
Atom: Number | Identified | [- sqrt] Atom | '(' Expr ')'
For parsing I'm using recursive descent parser. Let's say I'd like to parse this:
" 2 - 1 + 1 = 0"
and the result is 0, parser creates wrong tree:
-
/ \
2 +
/ \
1 1
How can I make this grammar left-associative? I'm newbie at this, please can you advice me source where I can find more information? Can I achieve this with recursive descent parser?
Take a look at Parsing Expressions by Recursive Descent by Theodore Norvell
There he gives three ways to solve the problem.
1. The shunting yard algorithm.
2. Classic solution by factoring the grammar.
3. Precedence climbing
Your problem stems from the fact that your grammar needs several changes, one example being
Exrp: Term { [+-] Expr} | Term
notice the addition of the { } around [+-] Expr indicating that they should be parsed together and that there can 0 or more of them.
Also by default you are building the tree as
-
/ \
2 +
/ \
1 1
i.e. -(2,+(1,1))
when you should be building the tree for left associative operators of the same precedence as
+
/ \
- 1
/ \
2 1
i.e. +(-(2,1),1)
Since there are three ways to do this in the paper I won't expand on them here. Also you mention that you are new to this so you should get a good compiler book to understand the details of you will encounter reading the paper. Most of these methods are implemented in common programming languages and available free on the internet, but be aware that many people do what you do and post wrong results.
The best way to check if you have it right is with a test like this using a multiple sequence of subtraction operations:
7-3-2 = 2
if you get
7-3-2 = 6 or something else
then it is wrong.
So I have been reading a bit on lexers, parser, interpreters and even compiling.
For a language I'm trying to implement I settled on a Recrusive Descent Parser. Since the original grammar of the language had left-recursion, I had to slightly rewrite it.
Here's a simplified version of the grammar I had (note that it's not any standard format grammar, but somewhat pseudo, I guess, it's how I found it in the documentation):
expr:
-----
expr + expr
expr - expr
expr * expr
expr / expr
( expr )
integer
identifier
To get rid of the left-recursion, I turned it into this (note the addition of the NOT operator):
expr:
-----
expr_term {+ expr}
expr_term {- expr}
expr_term {* expr}
expr_term {/ expr}
expr_term:
----------
! expr_term
( expr )
integer
identifier
And then go through my tokens using the following sub-routines (simplified pseudo-code-ish):
public string Expression()
{
string term = ExpressionTerm();
if (term != null)
{
while (PeekToken() == OperatorToken)
{
term += ReadToken() + Expression();
}
}
return term;
}
public string ExpressionTerm()
{
//PeekToken and ReadToken accordingly, otherwise return null
}
This works! The result after calling Expression is always equal to the input it was given.
This makes me wonder: If I would create AST nodes rather than a string in these subroutines, and evaluate the AST using an infix evaluator (which also keeps in mind associativity and precedence of operators, etcetera), won't I get the same result?
And if I do, then why are there so many topics covering "fixing left recursion, keeping in mind associativity and what not" when it's actually "dead simple" to solve or even a non-problem as it seems? Or is it really the structure of the resulting AST people are concerned about (rather than what it evaluates to)? Could anyone shed a light, I might be getting it all wrong as well, haha!
The shape of the AST is important, since a+(b*3) is not usually the same as (a+b)*3 and one might reasonably expect the parser to indicate which of those a+b*3 means.
Normally, the AST will actually delete parentheses. (A parse tree wouldn't, but an AST is expected to abstract away syntactic noise.) So the AST for a+(b*3) should look something like:
Sum
|
+---+---+
| |
Var Prod
| |
a +---+---+
| |
Var Const
| |
b 3
If you language obeys usual mathematical notation conventions, so will the AST for a+b*3.
An "infix evaluator" -- or what I imagine you're referring to -- is just another parser. So, yes, if you are happy to parse later, you don't have to parse now.
By the way, showing that you can put tokens back together in the order that you read them doesn't actually demonstrate much about the parser functioning. You could do that much more simply by just echoing the tokenizer's output.
The standard and easiest way to deal with expressions, mathematical or other, is with a rule hierarchy that reflects the intended associations and operator precedence:
expre = sum
sum = addend '+' sum | addend
addend = term '*' addend | term
term = '(' expre ')' | '-' integer | '+' integer | integer
Such grammars let the parse or abstract trees be directly evaluatable. You can expand the rule hierarchy to include power and bitwise operators, or make it part of the hierarchy for logical expressions with and or and comparisons.
I've got a simple grammar. Actually, the grammar I'm using is more complex, but this is the smallest subset that illustrates my question.
Expr ::= Value Suffix
| "(" Expr ")" Suffix
Suffix ::= "->" Expr
| "<-" Expr
| Expr
| epsilon
Value matches identifiers, strings, numbers, et cetera. The Suffix rule is there to eliminate left-recursion. This matches expressions such as:
a -> b (c -> (d) (e))
That is, a graph where a goes to both b and the result of (c -> (d) (e)), and c goes to d and e. I'm trying to produce an abstract syntax tree for these expressions, but I'm running into difficulty because all of the operators can accept any number of operands on each side. I'd rather keep the logic for producing the AST within the recursive descent parsing methods, since it avoids having to duplicate the logic of extracting an expression. My current strategy is as follows:
If a Value appears, push it to the output.
If a From or To appears:
Output a separator.
Get the next Expr.
Create a Link node.
Pop the first set of operands from output into the Link until a separator appears.
Erase the separator discovered.
Pop the second set of operands into the Link until a separator.
Push the Link to the output.
If I run this through without obeying steps 2.3–2.7, I get a list of values and separators. For the expression quoted above, a -> b (c -> (d) (e)), the output should be:
A sep_1 B sep_2 C sep_3 D E
Applying the To rule would then yield:
A sep_1 B sep_2 (link from C to {D, E})
And subsequently:
(link from A to {B, (link from C to {D, E})})
The important thing to note is that sep_2, crucial to delimit the left-hand operands of the second ->, does not appear, so the parser believes that the expression was actually written:
a -> (b c -> (d) (e))
In order to solve this with my current strategy, I would need a way to produce a separator between adjacent expressions, but only if the current expression is a From or To expression enclosed in parentheses. If that's possible, then I'm just not seeing it and the answer ought to be simple. If there's a better way to go about this, however, then please let me know!
I haven't tried to analyze it in detail, but: "From or To expression enclosed in parentheses" starts to sound a lot like "context dependent", which recursive descent can't handle directly. To avoid context dependence you'll probably need a separate production for a From or To in parentheses vs. a From or To without the parens.
Edit: Though it may be too late to do any good, if my understanding of what you want to match is correct, I think I'd write it more like this:
Graph :=
| List Sep Graph
;
Sep := "->"
| "<-"
;
List :=
| Value List
;
Value := Number
| Identifier
| String
| '(' Graph ')'
;
It's hard to be certain, but I think this should at least be close to matching (only) the inputs you want, and should make it reasonably easy to generate an AST that reflects the input correctly.
I'm trying to learn about shift-reduce parsing. Suppose we have the following grammar, using recursive rules that enforce order of operations, inspired by the ANSI C Yacc grammar:
S: A;
P
: NUMBER
| '(' S ')'
;
M
: P
| M '*' P
| M '/' P
;
A
: M
| A '+' M
| A '-' M
;
And we want to parse 1+2 using shift-reduce parsing. First, the 1 is shifted as a NUMBER. My question is, is it then reduced to P, then M, then A, then finally S? How does it know where to stop?
Suppose it does reduce all the way to S, then shifts '+'. We'd now have a stack containing:
S '+'
If we shift '2', the reductions might be:
S '+' NUMBER
S '+' P
S '+' M
S '+' A
S '+' S
Now, on either side of the last line, S could be P, M, A, or NUMBER, and it would still be valid in the sense that any combination would be a correct representation of the text. How does the parser "know" to make it
A '+' M
So that it can reduce the whole expression to A, then S? In other words, how does it know to stop reducing before shifting the next token? Is this a key difficulty in LR parser generation?
Edit: An addition to the question follows.
Now suppose we parse 1+2*3. Some shift/reduce operations are as follows:
Stack | Input | Operation
---------+-------+----------------------------------------------
| 1+2*3 |
NUMBER | +2*3 | Shift
A | +2*3 | Reduce (looking ahead, we know to stop at A)
A+ | 2*3 | Shift
A+NUMBER | *3 | Shift (looking ahead, we know to stop at M)
A+M | *3 | Reduce (looking ahead, we know to stop at M)
Is this correct (granted, it's not fully parsed yet)? Moreover, does lookahead by 1 symbol also tell us not to reduce A+M to A, as doing so would result in an inevitable syntax error after reading *3 ?
The problem you're describing is an issue with creating LR(0) parsers - that is, bottom-up parsers that don't do any lookahead to symbols beyond the current one they are parsing. The grammar you've described doesn't appear to be an LR(0) grammar, which is why you run into trouble when trying to parse it w/o lookahead. It does appear to be LR(1), however, so by looking 1 symbol ahead in the input you could easily determine whether to shift or reduce. In this case, an LR(1) parser would look ahead when it had the 1 on the stack, see that the next symbol is a +, and realize that it shouldn't reduce past A (since that is the only thing it could reduce to that would still match a rule with + in the second position).
An interesting property of LR grammars is that for any grammar which is LR(k) for k>1, it is possible to construct an LR(1) grammar which is equivalent. However, the same does not extend all the way down to LR(0) - there are many grammars which cannot be converted to LR(0).
See here for more details on LR(k)-ness:
http://en.wikipedia.org/wiki/LR_parser
I'm not exactly sure of the Yacc / Bison parsing algorithm and when it prefers shifting over reducing, however I know that Bison supports LR(1) parsing which means it has a lookahead token. This means that tokens aren't passed to the stack immediately. Rather they wait until no more reductions can happen. Then, if shifting the next token makes sense it applies that operation.
First of all, in your case, if you're evaluating 1 + 2, it will shift 1. It will reduce that token to an A because the '+' lookahead token indicates that its the only valid course. Since there are no more reductions, it will shift the '+' token onto the stack and hold 2 as the lookahead. It will shift the 2 and reduce to an M since A + M produces an A and the expression is complete.