BNF语法为一个简单的c ++程序示例 [英] BNF grammar for a simple c++ program example

问题描述

所以我试图为一个简单的c ++程序编写语法。

这是语法现在的样子：

  PDefs。 Program :: = [Def]; 
 terminator Def; 
 comment//; 
 comment/ ** /; 
 comment＃; 
 DFun。 Def :: = Type Id（[Arg]）{[Stm]}; 
 separator Arg，; 
 terminator Stm; 
 ADecl。 Arg :: = Type Id; 
 SExp。 Stm :: = Exp; ; 
 SDecl。 Stm :: = Type Id; ; 
 SDecls。 Stm :: = Type Id，[Id]; ; 
 SInit。 Stm :: = Type Id=Exp; ; 
 SReturn。 Stm :: =returnExp; ; 
 SWhile。 Stm :: =while（Exp）Stm; 
 SBlock。 Stm :: ={[Stm]}; 
 SIfElse。 Stm :: =if（Exp）StmelseStm; 
 
 
 EInt。 Exp15 :: = Integer; 
 EDouble。 Exp15 :: = Double; 
 ETrue。 Exp15 :: =true; 
 EFalse。 Exp15 :: =false; 
 EId。 Exp15 :: = Id; 
 EApp。 Exp15 :: = Id（[Exp]）; 
 EPIncr。 Exp14 :: = Exp15++; 
 EPDecr。 Exp14 :: = Exp15 - ; 
 EIncr。 Exp13 :: =++Exp14; 
 EDecr。 Exp13 :: = - Exp14; 
 ETimes。 Exp12 :: = Exp12*Exp13; 
 EDiv。 Exp12 :: = Exp12/Exp13; 
 EPlus。 Exp11 :: = Exp11+Exp12; 
 EMinus。 Exp11 :: = Exp11 - Exp12; 
 ELt。 Exp9 :: = Exp9< Exp10; 
 EGt。 Exp9 :: = Exp9> Exp10; 
 ELtEq。 Exp9 :: = Exp9< =Exp10; 
 EGtWq。 Exp9 :: = Exp9> =Exp10; 
 EEq。 Exp8 :: = Exp8==Exp9; 
 ENEq。 Exp8 :: = Exp8！=Exp9; 
 EAnd。 Exp4 :: = Exp4&& Exp5; 
 EOr。 Exp3 :: = Exp3|| Exp4; 
 EAss。 Exp2 :: = Exp3=Exp2; 
 
 coercions Exp 15; 
 separator Exp，; 
 separator Id，; 
 
 Tbool。 Type :: =bool; 
 Tdouble。 Type :: =double; 
 Tint。 Type :: =int; 
 Tvoid。 Type :: =void; 
 
令牌ID（字母（数字|数字|'_'）*）; 
  

这是需要解析的简单c ++程序

  //一个小的C ++程序
 #include< iostream> 
 
 int main（）
 {
 std :: cout< 你好，世界！ << std :: endl; 
 return 0; 
} 
  

所以当我试图解析它我得到错误在第6行意味着std :: cout行。因为我是新的bnf我不知道如何想解决这个。如果有人能给你一个例子，你将如何去解决这样的情况将是巨大的。

谢谢！

解决方案

它失败的行不能被解析，因为你缺少一些规则：

1. 您需要一个规则来解析合格的ID 。
   $ $ $
   限定的ID是一种特殊类型的标识符，可以（为您的目的）在与（不合格）标识符相同的情况下。
   
   std :: cout 和 std :: endl 是合格的id，并且它们的（简化）规则可以看起来像这样：

    < qualified_id> :: =< nested_name_specifier> < unqualified_id> 
   < nested_name_specifier> :: =< namespace_name> ::< nested_name_specifier> ;? < unqualified_id> 和   的 
     

   
   
   <




2. 您需要一个规则来解析表达式。 <$ c> $ c><< 运算符。
   
   此附加类型表达式的（简化）规则可能如下所示：

    < shift_left_expression> :: =< other_expression> 
   < shift_left_expression> :: =< shift_left_expression> << < other_expression> 
    

   其中（为您的目的）< other_expression>

。
字符串字面值（string literal）
是一种文字类型，可以作为表达式的一部分（例如标识符）使用（用于您的目的）。

Hello，world！是一个字符串文字，一个（简化的）规则可以看起来像这样：

 < string_literal> :: =\< s_char_sequence> ;?\
< s_char_sequence> :: =< s_char> 
< s_char_sequence> :: =< s_char_sequence> < s_char> 
  

其中< s_char> 是您要允许的任何字符在一个字符串字面量（为了保持简单，不允许字符，例如。）。






So i am trying to write grammar for a simple c++ program.

this is how the grammar looks like right now:

PDefs. Program ::= [Def] ;
terminator Def "" ;
comment "//" ;
comment "/*" "*/" ;
comment "#" ;
DFun. Def ::= Type Id "(" [Arg] ")" "{" [Stm] "}" ;
separator Arg "," ;
terminator Stm "" ;
ADecl. Arg ::= Type Id ;
SExp. Stm ::= Exp ";" ;
SDecl. Stm ::= Type Id ";" ;
SDecls. Stm ::= Type Id "," [Id] ";" ;
SInit. Stm ::= Type Id "=" Exp ";" ;
SReturn. Stm ::= "return" Exp ";" ;
SWhile. Stm ::= "while" "(" Exp ")" Stm ;
SBlock. Stm ::= "{" [Stm] "}" ;
SIfElse. Stm ::= "if" "(" Exp ")" Stm "else" Stm ;


EInt. Exp15 ::= Integer ;
EDouble. Exp15 ::= Double ;
ETrue. Exp15 ::= "true" ;
EFalse. Exp15 ::= "false" ;
EId. Exp15 ::= Id ;
EApp. Exp15 ::= Id "(" [Exp] ")" ;
EPIncr. Exp14 ::= Exp15 "++" ;
EPDecr. Exp14 ::= Exp15 "--" ;
EIncr. Exp13 ::= "++" Exp14 ;
EDecr. Exp13 ::= "--" Exp14 ;
ETimes. Exp12 ::= Exp12 "*" Exp13 ;
EDiv. Exp12 ::= Exp12 "/" Exp13 ;
EPlus. Exp11 ::= Exp11 "+" Exp12 ;
EMinus. Exp11 ::= Exp11 "-" Exp12 ;
ELt. Exp9 ::= Exp9 "<" Exp10 ;
EGt. Exp9 ::= Exp9 ">" Exp10 ;
ELtEq. Exp9 ::= Exp9 "<=" Exp10 ;
EGtWq. Exp9 ::= Exp9 ">=" Exp10 ;
EEq. Exp8 ::= Exp8 "==" Exp9 ;
ENEq. Exp8 ::= Exp8 "!=" Exp9 ;
EAnd. Exp4 ::= Exp4 "&&" Exp5 ;
EOr. Exp3 ::= Exp3 "||" Exp4 ;
EAss. Exp2 ::= Exp3 "=" Exp2 ;

coercions Exp 15 ;
separator Exp "," ;
separator Id "," ;

Tbool. Type ::= "bool" ;
Tdouble. Type ::= "double" ;
Tint. Type ::= "int" ;
Tvoid. Type ::= "void" ;

token Id (letter (letter | digit | '_')*) ;

and this is the simple c++ program that needs to be parsed

// a small C++ program
#include <iostream>

int main()
{
    std::cout << "Hello, world!" << std::endl;
    return 0;
}

so when i try to parse it i get the error in line 6 meaning the std::cout line. Since i am new to bnf i do not know how to "think" to solve this. If someone could give an example of how you would go to solve a situation like this would be great.!

Thank you!

解决方案

The line on which it fails, cannot be parsed because you are missing some rules :

1. You need a rule for parsing qualified ids.
   A qualified id is a special type of identifier, and can (for your purposes) be used in the same situations as an (unqualified) identifier.
   std::cout and std::endl are qualified ids, and a (simplified) rule for them could look something like this :

   <qualified_id> ::= <nested_name_specifier> <unqualified_id>
   <nested_name_specifier> ::= <namespace_name> "::" <nested_name_specifier>?
   

   in which (for your purposes), <unqualified_id> and <namespace_name> can be treated as identifiers.

2. You need a rule for parsing an expression with the << operator.
   A (simplified) rule for this additional type of expression could look something like this :

   <shift_left_expression> ::= <other_expression>
   <shift_left_expression> ::= <shift_left_expression> "<<" <other_expression>
   

   in which (for your purposes) <other_expression> stands for any other type of expression.

3. You need a rule for parsing string literals.
   A string literal is a type of literal, and it can be used (for your purposes) as part of an expression, like an identifier.
   "Hello, world!" is a string literal, and a (simplified) rule for them could look something like this :

   <string_literal> ::= "\"" <s_char_sequence>? "\""
   <s_char_sequence> ::= <s_char>
   <s_char_sequence> ::= <s_char_sequence> <s_char>
   

   in which <s_char> is any character that you want to allow inside a string literal (to keep it simple, don't allow the " character in there eg.).

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