how to resolve 2+2 and 2++2 conflict

问题

In larger program I have given the following (flex/bison)

In flex:

pn [\+|\-]
dig [0-9]+
exp [e|E]{dig}+

.

.

.

"+"             {printf("+ detected\n");
                return PLUS_SIGN;}




{pn}?{dig}+         { printf("digit detected - %s\n",yytext);
                sscanf(yytext, "%d", (int*)&yylval);
                return TYPE_INT;}

In Bison:

expr:
    expr PLUS_SIGN expr
      {
        $$ = $1 + $3;
        printf(" $$=%f\n",$$);
      }
    | TYPE_INT
      {     
        $$ = (int)$1;
        printf(" $$=%f\n",$$);
      }
;

The problem is:

When I give 2+2 it recognizes 2 and +2 instead of 2 , + , 2

How can I get it to do the addition?

回答1:

{pn}?{dig}+

Don't make the plus or minus sign ({pn?}) part of the number token. Treat them as two separate tokens, + and 2. Then flex won't have any ambiguity to resolve.

{dig}+

Instead, have bison handle the unary plus and minus operators. Make it the parser's job, not the lexer's.

| PLUS_SIGN expr
  {
    $$ = +$2;
    printf(" $$=%f\n",$$);
  }
| MINUS_SIGN expr
  {
    $$ = -$2;
    printf(" $$=%f\n",$$);
  }

回答2:

The grammar shows the left part and the right part of PLUS_SIGN has same priority when reducing symbol.The PLUS_SIGN is left combination, so the new grammar is below:

expr: expr PLUS_SIGN expr2
      {
         $$ = $1 + $3;
         printf("$$=%f\n", $$);
      }
    | expr2
     {
        $$ = $1;
     }
;
expr2: TYPE_INT
     {     
        $$ = (int)$1;
        printf(" $$=%f\n",$$);
     }
;

来源：https://stackoverflow.com/questions/12540757/how-to-resolve-22-and-22-conflict