CODEWITHSUNDEEP
compiler-constructionbison
Vito
Vito

Reputation: 65

$0 of has no declared type

Bison compiler returns me the error $0 of 'routine' has no declared type. I'm trying to print a message declaring a fonction, so I have to print the opening brace before printing the body of the function. I read the doc, and I found a solution coming from that page :Bison 3.0.4: Actions so I tried to apply it to my problem :

fun_decl : 
    type routine fun {end_fonction();};

routine :
    {declare_fonction($0);};

I also specified a type for the non terminal 'type' :

%type <type_object> type

So I don't understand where does my mistake come from.

NB : When I simply associate an empty instruction on the non terminal routine, bison also returns a mistake : "rule useless in parser due to conflicts". Does it mean I can't have multiple empty rules ?

Upvotes: 1

Views: 970

Answers (3)

user207421
user207421

Reputation: 310980

You haven't declared the type of routine:

%type <type_object> routine

Upvotes: 0

Vito
Vito

Reputation: 65

I just found the solution to my problem, I had to specify the type of $0, to let Bison know the size to take in the stack. In my case :

routine :
    {declare_fonction($<type_object>0);};

Upvotes: 1

rici
rici

Reputation: 241861

You can just use a mid-rule action:

fun_decl : 
    type {declare_fonction($1);} fun {end_fonction();};

This has the advantage that bison knows the type of $1. It never tries to figure out the type of $0 because it is not possible in the general case.

As discussed in the manual, mid-rule actions can create shift-reduce conflicts. "Marker" (empty) non-terminals, like your routine, can also create shift-reduce conflicts, and roughly under the same circumstances. But that will be a different question.

Upvotes: 2

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