【问题标题】:unrecognized rule in lexlex中无法识别的规则
【发布时间】:2010-04-25 09:57:12
【问题描述】:

我正在用 lex 编写一个程序,它给了我以下错误:

scanner.l:49: 无法识别的规则

第 49 行是:{number} {return(NUM);}

编辑: 但是,该错误似乎与之前的行有关,{id} {return(ID);}。 它将直接在该规则之后列出作为错误来源的行,即使它是空白的。

这是我的代码:

#include <stdio.h>

%token BOOL, ELSE, IF, TRUE, WHILE, DO, FALSE, INT, VOID
%token LPAREN, RPAREN, LBRACK, RBRACK, LBRACE, RBRACE, SEMI, COMMA, PLUS, MINUS, TIMES
%token DIV, MOD, AND, OR, NOT, IS, ADDR, EQ, NE, LT, GT, LE, GE
%token NUM, ID, PUNCT, OP

int line = 1, numAttr;
char *strAttr;
%}

/* regular definitions */

delim   [ \t]
ws      {delim}+
letter  [A-Za-z]
digit   [0-9]
id      ({letter} | _)({letter} | {digit} | _)*
number  {digit}+

%%

{ws}        {/* no action and no return */}
[\n]        {line++;}
bool        {return(BOOL);}
else        {return(ELSE);}
if          {return(IF);}
true        {return(TRUE);}
while       {return(WHILE);}
do          {return(DO);}
false       {return(FALSE);}
int         {return(INT);}
void        {return(VOID);}

{id}        {return(ID);}  // error is on these two lines
{number}    {return(NUM);} //

"("         {yylval = LPAREN; return(PUNCT);}
")"         {yylval = RPAREN; return(PUNCT);}
"["         {yylval = LBRACK; return(PUNCT);}
"]"         {yylval = RBRACK; return(PUNCT);}
"{"         {yylval = LBRACE; return(PUNCT);}
"}"         {yylval = RBRACE; return(PUNCT);}
";"         {yylval = SEMI;   return(PUNCT);}
","         {yylval = COMMA;  return(PUNCT);}

"+"         {yylval = PLUS;   return(OP);}
"-"         {yylval = MINUS;  return(OP);}
"*"         {yylval = TIMES;  return(OP);}
"/"         {yylval = DIV;    return(OP);}
"%"         {yylval = MOD;    return(OP);}
"&"         {yylval = ADDR;   return(OP);}
"&&"        {yylval = AND;    return(OP);}
"||"        {yylval = OR;     return(OP);}
"!"         {yylval = NOT;    return(OP);}
"!="        {yylval = NE;     return(OP);}
"="         {yylval = IS;     return(OP);}
"=="        {yylval = EQ;     return(OP);}
"<"         {yylval = LT;     return(OP);}
"<="        {yylval = LE;     return(OP);}
">"         {yylval = GT;     return(OP);}
">="        {yylval = GE;     return(OP);}

%%

这条规则有什么问题?谢谢。

【问题讨论】:

    标签: lex


    【解决方案1】:

    上一行导致了问题。如果在{id} 规则和{number} 规则之间添加一个空格,您会注意到错误的行号没有改变。

    模式中不允许有空格。如此定义 {id}:

    id      ({letter}|_)({letter}|{digit}|_)*
    

    【讨论】:

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