EBNF 语法到ANTLR3? [英] EBNF grammar to ANTLR3?

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本文介绍了EBNF 语法到ANTLR3?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有 Jass 脚本语言的 EBNF 语法.
需要做什么才能将其转换为与 ANTLR 3.5 一起使用?
此外,是否有任何工具可以帮助我这样做?

I have this EBNF grammar for the Jass scripting language.
What needs to be done to convert it to work with ANTLR 3.5?
Furthermore, are there any sort of tools available to aid me in doing so?

//----------------------------------------------------------------------
// Global Declarations
//----------------------------------------------------------------------
program  ::= file+
file     ::= newline? ( declr newline )* func*
declr    ::= typedef
           | globals
           | native_func
typedef  ::= 'type' id 'extends' ( 'handle' | id )
globals  ::= 'globals' newline global_var_list 'endglobals'
global_var_list
         ::= ( 'constant' type id '=' expr newline | var_declr newline )*
native_func
         ::= 'constant'? 'native' func_declr
func_declr
         ::= id 'takes' ( 'nothing' | param_list ) 'returns' ( type | 'nothing' )
param_list
         ::= type id ( ',' type id )*
func     ::= 'constant'? 'function' func_declr newline local_var_list statement_list 'endfunction' newline

//----------------------------------------------------------------------
// Local Declarations
//----------------------------------------------------------------------
local_var_list
         ::= ( 'local' var_declr newline )*
var_declr
         ::= type id ( '=' expr )?
           | type 'array' id
statement_list
         ::= ( statement newline )*
statement
         ::= set
           | call
           | ifthenelse
           | loop
           | exitwhen
           | return
           | debug
set      ::= 'set' id '=' expr
           | 'set' id '[' expr ']' '=' expr
call     ::= 'call' id '(' args? ')'
args     ::= expr ( ',' expr )*
ifthenelse
         ::= 'if' expr 'then' newline statement_list else_clause? 'endif'
else_clause
         ::= 'else' newline statement_list
           | 'elseif' expr 'then' newline statement_list else_clause?
loop     ::= 'loop' newline statement_list 'endloop'
exitwhen ::= 'exitwhen' expr
return   ::= 'return' expr?
debug    ::= 'debug' ( set | call | ifthenelse | loop )

//----------------------------------------------------------------------
// Expressions
//----------------------------------------------------------------------
expr     ::= binary_op
           | unary_op
           | func_call
           | array_ref
           | func_ref
           | id
           | const
           | parens
binary_op
         ::= expr ( [+-*/><] | '==' | '!=' | '>=' | '<=' | 'and' | 'or' ) expr
unary_op ::= ( '+' | '-' | 'not' ) expr
func_call
         ::= id '(' args? ')'
array_ref
         ::= id '[' expr ']'
func_ref ::= 'function' id
const    ::= int_const
           | real_const
           | bool_const
           | string_const
           | 'null'
int_const
         ::= decimal
           | octal
           | hex
           | fourcc
decimal  ::= [1-9] [0-9]*
octal    ::= '0' [0-7]*
hex      ::= '$' [0-9a-fA-F]+
           | '0' [xX] [0-9a-fA-F]+
fourcc   ::= '' ' .{4} ' ''
real_const
         ::= [0-9]+ '.' [0-9]*
           | '.' [0-9]+
bool_const
         ::= 'true'
           | 'false'
string_const
         ::= '"' .* '"'
parens   ::= '(' expr ')'

//----------------------------------------------------------------------
// Base RegEx
//----------------------------------------------------------------------
type     ::= id
           | 'code'
           | 'handle'
           | 'integer'
           | 'real'
           | 'boolean'
           | 'string'
id       ::= [a-zA-Z] ( [a-zA-Z0-9_]* [a-zA-Z0-9] )?
newline  ::= '\n'+


在此先感谢您提供的任何建议!


Thanks in advance to any advice offered!

推荐答案

免责声明:我实际上并不使用 ANTLR,因此使用的人可能会提供更详细的信息.

Disclaimer: I don't actually use ANTLR, so someone that does might come by with more detailed information.

ANTLR 生成递归下降解析器,因此必须重构您的语法以消除左递归,例如在 expr 中:

ANTLR generates recursive descent parsers, so your grammar will have to be refactored to eliminate left recursion, which you have e.g. in expr:

expr     ::= binary_op
...
binary_op
         ::= expr ( [+-*/><] | '==' | '!=' | '>=' | '<=' | 'and' | 'or' ) expr

在解析 expr 时,解析器会尝试将 binary_op 作为一个选项,遇到另一个 expr,然后尝试递归解析它而不消耗任何输入,你现在都会有无限递归.

While parsing expr, the parser would try binary_op as an option, encounter another expr, then try to parse that recursively without having consumed any input, and you would now have infinite recursion.

这通常是通过按照

expr     ::= binary_op
...
binary_op
         ::= term ( [+-] term )

term = factor ( [*/] factor)

factor = id
         | const
         | parens
         ...

等等.

不是一个微不足道的过程,但也不是不可能做到.

Not a trivial process, but not impossible to do either.

这篇关于EBNF 语法到ANTLR3?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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