MGrammar Equivalent of Yacc's $$?

Question

Consider the following simple calculator implemeted in yacc:

1	%{
2	#include <stdio.h>
3	int yylex(void);
4	void yyerror(char *);
5	%}
6
7	%token INTEGER
8
9	%%
10
11	program:
12	program expr '\n'         { printf("%d\n", $2); }
13	|
14	;
15
16	expr:
17	INTEGER
18	| expr '+' expr           { $$ = $1 + $3; }
19	| expr '-' expr           { $$ = $1 - $3; }
20	;
21
22	%%
23
24	void yyerror(char *s) {
25	fprintf(stderr, "%s\n", s);
26	}
27
28	int main(void) {
29	yyparse();
30	return 0;
31	}
32

I am interested in how to handle yacc's $$ construct in MGrammar. As you can see in lines 18 and 19 above, the $$ holds the result of an expression calculated in the "codebehind" (for lack of a better term) for the expression. This allows the result to be used in parse trees as if it were brought in as a literal.

How would I do the same in MGrammar? I am familiar with navigating parse trees in .NET, so, if you say "you need C# for that", that's fine ... in fact, I almost expect that answer. I know how to calculate the result in the # code, but where do I return the result of the expression to be used in the parse tree?

Thank you,
Jeff Ferguson

Answer 1

Hi Jeff,

 There's no directequivalent in MGrammar for Yacc's $$ macro, but there isn't a 1:1 relationship between these two technologies
 either. MGrammar does both the work that Lex and Yacc solve together - and you can easily output nodes to the AST that will
 do the same thing for you, e.g.:

 syntax SomeSyntax = a:tNumber op:tOperator b:tNumber => Add { Left { a }, Right { b} };

 In the backend you will have to walk the AST and take actions accordingly.

 HTH,

 --larsw

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Lars Wilhelmsen | Senior Consultant | Miles, Norway | Connected Systems MVP | http://larswilhelmsen.com/

Answer 2

Thanks, Lars ... as I said in my original question, "you need C# for that" is an acceptable answer. I just thought that, if there were some equivalent that I was missing, I wanted to find out what it was.

Answer 3

Hi jeff,
I have implemented a grammar that lets me parse various binary expressions in MGrammar I could share with you if you're interested. I found it pretty hard to use MGrammar to parse epxressions and maintain precedence but I managed to figure it out.

Where:
x + y => (x + y)
x + y * z => x + (y * z)
x + y and z | w => x + (y and (z | w))

For example. Very tricky.

Answer 4

Hey there! Sure, I'd love to se it. Thanks!

Answer 5

Hi again,

 I assume that the problem you had with the expressions in your post is related to associativity and precedence. In
 MGrammar you can use precedence, left(n) and right(n) to solve this.

 For the basic mathematical operators, multiply and divide have higher precedence than plus and minus.

 A small example of the use of left & right:

 syntax tOperator = left(1) tPlus => "Plus"
                         | left(1) tMinus => "Minus"
                         | left(2) tDivide => "Divide"
                         | left(2) tMultiply => "Multiply"
                         | right(3) tExponentiation => "Exp";
        token tPlus = "+";
        token tMinus = "-";
        token tDivide = "/";
        token tMultiply = "*";
        token tExponentiation = "**";

 The precedence keyword will help you solve the classical "danging else" problem. A more thorough explanation of this
 is done in the MGrammar specification that is shipped with the Oslo SDK.

 HTH,

 --larsw

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Lars Wilhelmsen | Senior Consultant | Miles, Norway | Connected Systems MVP | http://larswilhelmsen.com/

Answer 6

I'm just catching belatedly up on forum threads, but I thought I'd interject that we're considering ways to evaluate expressions directly in grammars. It's a pretty useful feature, and we'd love to provide the full M expression capabilities on the RHS of a grammar production. So stay tuned!

Paul [MSFT]

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Paul Vick MSFT

Answer 7

Hi Paul,

 That would be awesome! :-)

 --larsw

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Lars Wilhelmsen | Senior Consultant | Miles, Norway | Connected Systems MVP | http://larswilhelmsen.com/