CODEWITHSUNDEEP
parsingyacc
neuromancer
neuromancer

Reputation: 55539

In yacc, how to put a definition into a variable?

In my yacc file I have the following code:

fun_declaration : type_specifier ID '(' params ')' 
                  {$2->type = "function";
                   $2->args = params; }

params : param_list | VOID ;

Do you see what I'm trying to do?

args is a string. I'm trying to put the function parameters into this string. How to do that?

Upvotes: 0

Views: 147

Answers (2)

Chris Dodd
Chris Dodd

Reputation: 2960

You need to have 'params' return the string you want in $$, much like ID is returning a pointer to some struct with 'type' and 'args' fields. This means you'll need a %type declaration for it saying which element of the %union to use.

There are lots of books and online tutorials for how to use yacc like this.

Upvotes: 1

laalto
laalto

Reputation: 152867

Just refer to with $n it as you would refer to any semantic value component in the rule. Something like this:

 $2->args = strdup($4);

Upvotes: 0

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