CODEWITHSUNDEEP
bison
yhab
yhab

Reputation: 3

I get this message'rule useless in grammar'

I'm new to Bison. I wrote some rules but I get a lot of 'rule useless in grammar'.

I think 'expr' cause this problem. Please help me.

image of result

File : 
| class {printf("accepted");}
; 
class : CLASS IDENT '{' function  '}'
| CLASS IDENT '{'  global '}'
;
global : data_type IDENT 
;
function : 
|data_type IDENT'('Params')' '{'statement'}'
|VOID IDENT'('Params')' '{'statement'}'
;
Params : data_type IDENT 
| data_type IDENT',' Params
;
data_type : INT_T 
|DOUBLE_T
|BOOL_T 
|VOID;
;       
statement : WHILE '(' expr')'  statement  
| FOR'('data_type IDENT '=' expr ';' expr ';' expr')'  statement 
;   
expr:   expr COMP expr
|expr '=' expr
|INT_T
|BOOL_T     
;

Upvotes: 0

Views: 83

Answers (1)

rici
rici

Reputation: 241691

The problem is that statement has only recursive rules. That makes statement impossible to use in a derivation because any derivation starting with statement never terminates.

Because statement cannot be used to parse any finite string, bison removes it and all its rules from the grammar, as well as any rules which use statement. With those rules removed, expr and Params are no longer referred to in the grammar, so they are also marked useless.

You probably intended statement to have other, non-recursive alternatives, such as expr.

Upvotes: 1

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