简单的YCAA

%{
int tempCount1 = -1;
int tempCount2 = -1;
int tempCount3 = -1;
int tempCount4 = -1;
int tempCount5 = -1;
%}
%token NUMBER
%left '+' '-'
%left '*' '/'
%%
list:
|list 'q' {exit(0);}
|list '/n'
|list expr '/n' {printf("/t%ld/n", $2);}
;
expr: NUMBER { $$ = $1;}
|expr '+' expr {tempCount1= $1 + $3;$$ = $1 + $3;}
|expr '-' expr {tempCount2= $1 - $3;$$ = $1 - $3;}
|expr '*' expr {tempCount3= $1 * $3;$$ = $1 * $3;}
|expr '/' expr {tempCount4= $1 / $3;$$ = $1 / $3;}
|'(' expr ')' {tempCount5= $2;$$ = $2;}
;
%%
#include <stdio.h>
#include <ctype.h>
int lineno;
void main(int argc, char *argv[])
{
yyparse();
}
int yylex()
{
int c;
while((c = getchar()) == ' ' || c == '/t')
;
if(c == EOF)
return 0;
if(isdigit(c))
{
ungetc(c, stdin);
scanf("%d", &yylval);
return NUMBER;
}
if(c == '/n')
lineno++;
return c;
}
yyerror(char *s)
{
fprintf(stderr, "%s", s);
fprintf(stderr, " near line %d/n", lineno);
}

--------------------------------------------------------------------------------------

%left '+' '-'
%left '*' '/'

表示其后的算符是遵循左结合的

当我入力2*(3-1)的时候,NUMBER先分别找到2,3,1.....(NUMBER)

然后tempCount2 = 3-1($$ = $1 - $3;)

tempCount5 = 2            ($$ = $2)

tempCount3 = 4           ($$ = $1 * $3)