Question

I am learning bison through the standard calculator example, where I am trying to add some new functionalities to already written calculator. I am trying to add a feature to my calc where a+1 = b would work.

in my .y file I have to declare terminal symbols :

%token <symbol> SYMBOL_NAME
%token <dval> NUMBER
%token LIST_VAR_COMMAND
%token LIST_CONST_COMMAND
%token LIST_FUNC_COMMAND
%token DELETE_COMMAND
%token QUIT_COMMAND
%token CLEAR_COMMAND
%token BINARY
%token OCTAL
%token DECIMAL
%token HEXADECIMAL
%token UNSIGNED
%token CHAR
%token SHORT
%token INT
%token LONG
%token FOR

I declare my statement to be like :

statement:      ';'
|       expression             { setp($1); print_universal_base($1, 0); }
|       expression BINARY      { setp($1); print_universal_base($1, 2); }
|       expression OCTAL       { setp($1); print_universal_base($1, 8); }
|       expression DECIMAL     { setp($1); print_universal_base($1, 10); }
|       expression HEXADECIMAL { setp($1); print_universal_base($1, 16); }

I declare my expression to be like :

expression:     expression ',' expression { $$ = $3; }
|       NUMBER { $$ = $1; }
|       CHAR { $$ =(int)$1; }

I get an error in line

|       CHAR { $$ =(int)$1; }

saying : $1 of `expression’ has no declared type

what am I doing wrong here?

Thanks.

Answer 1

To access the value of some symbol from the rule, that symbol needs to have a type, and its value needs to be set properly. So for the rule

expression: CHAR { $$ = $1; }

to work, CHAR needs to have a defined type, for example:

%token<dval> CHAR

and the lexer needs to provide the value in yylval.dval when it returns a CHAR token. Now depending on what you want the input a+1=b to actually do, (and assuming the lexer returns CHAR for a and b and not SYMBOL) this may or may not be what you want.