Hi Arjen,
Thanks for the mail...I found it very helpful and i started to execute
it..I succeeded in doing it...
However i started writing a sample program for converting "Infix to
Prefix"...I am struck at this point ...so can u help me in
writing this program...
Thanks in advance
Regards,
Arjun
On Sat, Jun 28, 2008 at 12:54 AM, Arjen Markus <[EMAIL PROTECTED]>
wrote:
> > Hi,
> > Can u send me any documentation on LEMON that u have worked out.
> > I have some queries also.
> > 1.Does LEMON work on Windows environment?
> > 2.I tried but it is not.I heard that it works on Linux environment.I am
> > trying to know why not on windows...can u give me some info about it
> >
> > I am reading the material u have suggested me but it did not have any
> > information regarding to my queries.
> >
> > Can u mail me if u have made any documentation on LEMON.
> >
>
> Well, basically lemon works as follows:
> 1. It reads the definition of the parser from the input file
> you give it as the first argument. From that it produces
> a set of tables that implement the parser. This information
> gets filled in in a template.
> 2. The result is a C file that you can use to parse (interpret)
> your input.
> 3. What it does not do is split up the input into tokens. You
> have to do that yourself.
>
> I attach a small - not very efficient - example that shows a
> few features of lemon.
>
> As for your questions:
> There is no reason it would not work on Windows: it is a
> perfectly straghtforward C program, you get the source code
> and compile it. One thing though: the template file must be
> in the directory where you run the program, where the
> executable lives or in the PATH.
>
> (I am using it myself on Windows :). I am adapting it so that
> it will produce Fortran code instead of C code. But that is
> off topic.)
>
> Here is my sample parser definition:
>
> // expr.y
> // Simple parser:
> // expr = vaue + value + value + ...
>
> %extra_argument {State *state}
> %token_type {Token*}
> %type term {int}
>
> expr ::= firstterm plusterms . {
>printf( "Result: %d\n", state->sum );
> }
>
> firstterm ::= term(T) . {
>state->sum = T;
>printf("First term: %d\n", T );
> }
>
> term(V) ::= NUMBER(N) . {
>sscanf( N->token, "%d", );
>printf("Term: %d -- %s\n", V, N->token );
> }
>
> plusterms ::= .
> plusterms ::= plusterms plusterm .
>
> plusterm ::= PLUS term(T) . {
>state->sum = state->sum + T;
>printf( "Result so far: %d\n", state->sum );
> }
>
> --
> And here is the main program that includes the
> resulting C code:
>
> #include
> #include
> #include
>
> typedef struct {
>char *token;
> } Token;
> typedef struct {
>int sum;
> } State;
>
> #include "expr.h"
> #include "expr.c"
>
> int main( int argc, char *argv[] ) {
>Token token[10];
>State state;
>
>void *pParser = (void *) ParseAlloc( malloc );
>
>ParseTrace( stdout, ">>" );
>
>token[0].token = "1"; Parse( pParser, NUMBER, [0], );
>token[1].token = "+"; Parse( pParser, PLUS, [1], );
>token[2].token = "2"; Parse( pParser, NUMBER, [2], );
>token[3].token = "+"; Parse( pParser, PLUS, [3], );
>token[4].token = "3"; Parse( pParser, NUMBER, [4], );
>
>Parse( pParser, 0, [0], );
>
>ParseFree( pParser, free );
>
>return 0;
> }
>
> -
> One thing to note:
> As the tokens are not always treated the moment they are passed
> to the parser - this depends on the grammar and the moment in the
> parsing process, you need to make sure that the tokens remain
> available. I have done that in this silly program by using an
> array of them. But a smarter program would use lemon's
> abilities to destroy the tokens when ready.
>
> Regards,
>
> Arjen
>
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