Question

我无法找出我的计划无效的原因。没有＆＃34; a = 5;＆＃34;在解析程序（这是下面的代码块）

%{
#include <iostream>
int line_count = 0;
#include "translator.h"
%}

%union 
{
Stmt *separator;
Expr *exp;
Stmt *statement;

int int_val;
char name[256];
};

%token  <int_val> NUM
%token <name> ID
%token STRING ENDLINE BOOLEAN INT MAIN PUBLIC CLASS VOID STATIC PRINTLN EQ
%start programm

%type <separator> separator
%type <statement> statement
%type <statement> assign_statement
%type <exp> exp
%type <exp> var_ref
%type <exp> literal_exp

%left EQ

%%
programm:
class_declaration
;

class_declaration:
PUBLIC CLASS ID opencurlybracket main_method_declaration closecurlybracket
;

main_method_declaration:
PUBLIC STATIC VOID MAIN '(' STRING '[' ']' ID ')' opencurlybracket method_body           closecurlybracket
;

method_body:
local_declarations
|
statements
;

statements:
statement
|
statements statement
;

statement:
assign_statement
;

assign_statement:
var_ref '=' exp ';' separator { $$ = new StmtAssign($1, $3); }
;

local_declarations:
local_declaration
|
local_declarations local_declaration
;

local_declaration:
type ID ';' separator
;

exp:
var_ref
|
literal_exp
;

var_ref:
ID { $$ = new ExprVar($1); }
;

literal_exp:
NUM { $$ = new ExprNum($1); }
;

type:
INT
|
BOOLEAN
;

separator:
ENDLINE
{
    ++line_count;
    $$ = new StmtEndline(line_count);   
}
;

opencurlybracket:
'{'
|
'{' separator
;

closecurlybracket:
'}'
|
'}' separator
;

%%

通过这种语法解析的代码：

public class Summ {
    public static void main(String[] args) {
        int a; 
        a = 5;
    }
}

和lex文件：

%{
#include <stdlib.h>
#include <stdio.h>
#include "translator.h"
%}

%option noyywrap

%%

(public)    { return (PUBLIC); }
(class)     { return (CLASS); }
(static)    { return (STATIC); }
(void)      { return (VOID); }
(main)      { return (MAIN); }
(String)    { return (STRING); }
(int)       { return (INT); }
(println)   { return (PRINTLN); }
(\{)        { return '{'; }
(\})        { return '}'; }
(\()        { return '('; }
(\))        { return ')'; }         
(\[)        { return '['; }
(\])        { return ']'; }     
(\;)        { return ';'; }
(\=)        { return '='; }
"=="        { return (EQ); }
(\n)        { return (ENDLINE); }
[a-zA-Z_\$][a-zA-Z0-9_\$]*  {
                printf("found %s\n", yytext);
                strcpy(yylval.name, yytext);
                return ID;  
            }
([0])|([1-9][0-9]*)     {
                printf("found %s\n", yytext);
                yylval.int_val = atoi(yytext);
                return NUM;
            }
" "     { /*space - do nothing*/ }
.       { /*do nothing*/ }
%%

我希望有人在我的代码中发现错误。

Answer 1

以下是您对method_body的定义：

method_body:
local_declarations
|
statements
;

这指定method_body包含local_declarations或statements，但不包含。您需要将其更改为接受local_declarations后跟statements。

yacc程序中的语法错误

1 个答案: