ANTLR解析语法->;树语法
编译器理论类的最后一个任务是为Java的一小部分(不是MiniJava)创建编译器。我们的教授给了我们使用任何我们想要的工具的选择,在反复摸索之后,我决定使用ANTLR。我已经设法让扫描器和解析器启动并运行,解析器输出AST。我现在一直在试图编译一个树语法文件。我知道基本的想法是从解析器中复制语法规则,并删除大部分代码,保留重写规则,但它似乎不想编译(犯了Token错误)。我走对了吗?我错过了一些琐碎的事情吗 树语法:ANTLR解析语法->;树语法,antlr,tree-grammar,Antlr,Tree Grammar,编译器理论类的最后一个任务是为Java的一小部分(不是MiniJava)创建编译器。我们的教授给了我们使用任何我们想要的工具的选择,在反复摸索之后,我决定使用ANTLR。我已经设法让扫描器和解析器启动并运行,解析器输出AST。我现在一直在试图编译一个树语法文件。我知道基本的想法是从解析器中复制语法规则,并删除大部分代码,保留重写规则,但它似乎不想编译(犯了Token错误)。我走对了吗?我错过了一些琐碎的事情吗 树语法: tree grammar J0_SemanticAnalysis; opt
tree grammar J0_SemanticAnalysis;
options {
language = Java;
tokenVocab = J0_Parser;
ASTLabelType = CommonTree;
}
@header
{
package ritterre.a4;
import java.util.Map;
import java.util.HashMap;
}
@members
{
}
walk
: compilationunit
;
compilationunit
: ^(UNIT importdeclaration* classdeclaration*)
;
importdeclaration
: ^(IMP_DEC IDENTIFIER+)
;
classdeclaration
: ^(CLASS IDENTIFIER ^(EXTENDS IDENTIFIER)? fielddeclaration* methoddeclaration*)
;
fielddeclaration
: ^(FIELD_DEC IDENTIFIER type visibility? STATIC?)
;
methoddeclaration
: ^(METHOD_DEC IDENTIFIER type visibility? STATIC? ^(PARAMS parameter+)? body)
;
visibility
: PRIVATE
| PUBLIC
;
parameter
: ^(PARAM IDENTIFIER type)
;
body
: ^(BODY ^(DECLARATIONS localdeclaration*) ^(STATEMENTS statement*))
;
localdeclaration
: ^(DECLARATION type IDENTIFIER)
;
statement
: assignment
| ifstatement
| whilestatement
| returnstatement
| callstatement
| printstatement
| block
;
assignment
: ^(ASSIGN IDENTIFIER+ expression? expression)
;
ifstatement
: ^(IF relation statement ^(ELSE statement)?)
;
whilestatement
: ^(WHILE relation statement)
;
returnstatement
: ^(RETURN expression?)
;
callstatement
: ^(CALL IDENTIFIER+ expression+)
;
printstatement
: ^(PRINT expression)
;
block
: ^(STATEMENTS statement*)
;
relation
// : expression (LTHAN | GTHAN | EQEQ | NEQ)^ expression
: ^(LTHAN expression expression)
| ^(GTHAN expression expression)
| ^(EQEQ expression expression)
| ^(NEQ expression expression)
;
expression
// : (PLUS | MINUS)? term ((PLUS | MINUS)^ term)*
: ^(PLUS term term)
| ^(MINUS term term)
;
term
// : factor ((MULT | DIV)^ factor)*
: ^(MULT factor factor)
| ^(DIV factor factor)
;
factor
: NUMBER
| IDENTIFIER (DOT IDENTIFIER | LBRAC expression RBRAC)?
| NULL
| NEW IDENTIFIER LPAREN RPAREN
| NEW (INT | IDENTIFIER) (LBRAC RBRAC)?
;
type
: (INT | IDENTIFIER) (LBRAC RBRAC)?
| VOID
;
语法分析器:
parser grammar J0_Parser;
options
{
output = AST; // Output an AST
tokenVocab = J0_Scanner; // Pull Tokens from Scanner
//greedy = true; // forcing this throughout?! success!!
//cannot force greedy true throughout. bad things happen and the parser doesnt build
}
tokens
{
UNIT;
IMP_DEC;
FIELD_DEC;
METHOD_DEC;
PARAMS;
PARAM;
BODY;
DECLARATIONS;
STATEMENTS;
DECLARATION;
ASSIGN;
CALL;
}
@header { package ritterre.a4; }
// J0 - Extended Specification - EBNF
parse
: compilationunit EOF -> compilationunit
;
compilationunit
: importdeclaration* classdeclaration*
-> ^(UNIT importdeclaration* classdeclaration*)
;
importdeclaration
: IMPORT IDENTIFIER (DOT IDENTIFIER)* SCOLON
-> ^(IMP_DEC IDENTIFIER+)
;
classdeclaration
: (PUBLIC)? CLASS n=IDENTIFIER (EXTENDS e=IDENTIFIER)? LBRAK (fielddeclaration|methoddeclaration)* RBRAK
-> ^(CLASS $n ^(EXTENDS $e)? fielddeclaration* methoddeclaration*)
;
fielddeclaration
: visibility? STATIC? type IDENTIFIER SCOLON
-> ^(FIELD_DEC IDENTIFIER type visibility? STATIC?)
;
methoddeclaration
: visibility? STATIC? type IDENTIFIER LPAREN (parameter (COMMA parameter)*)? RPAREN body
-> ^(METHOD_DEC IDENTIFIER type visibility? STATIC? ^(PARAMS parameter+)? body)
;
visibility
: PRIVATE
| PUBLIC
;
parameter
: type IDENTIFIER
-> ^(PARAM IDENTIFIER type)
;
body
: LBRAK localdeclaration* statement* RBRAK
-> ^(BODY ^(DECLARATIONS localdeclaration*) ^(STATEMENTS statement*))
;
localdeclaration
: type IDENTIFIER SCOLON
-> ^(DECLARATION type IDENTIFIER)
;
statement
: assignment
| ifstatement
| whilestatement
| returnstatement
| callstatement
| printstatement
| block
;
assignment
: IDENTIFIER (DOT IDENTIFIER | LBRAC a=expression RBRAC)? EQ b=expression SCOLON
-> ^(ASSIGN IDENTIFIER+ $a? $b)
;
ifstatement
: IF LPAREN relation RPAREN statement (options {greedy=true;} : ELSE statement)?
-> ^(IF relation statement ^(ELSE statement)?)
;
whilestatement
: WHILE LPAREN relation RPAREN statement
-> ^(WHILE relation statement)
;
returnstatement
: RETURN expression? SCOLON
-> ^(RETURN expression?)
;
callstatement
: IDENTIFIER (DOT IDENTIFIER)? LPAREN (expression (COMMA expression)*)? RPAREN SCOLON
-> ^(CALL IDENTIFIER+ expression+)
;
printstatement
: PRINT LPAREN expression RPAREN SCOLON
-> ^(PRINT expression)
;
block
: LBRAK statement* RBRAK
-> ^(STATEMENTS statement*)
;
relation
: expression (LTHAN | GTHAN | EQEQ | NEQ)^ expression
;
expression
: (PLUS | MINUS)? term ((PLUS | MINUS)^ term)*
;
term
: factor ((MULT | DIV)^ factor)*
;
factor
: NUMBER
| IDENTIFIER (DOT IDENTIFIER | LBRAC expression RBRAC)?
| NULL
| NEW IDENTIFIER LPAREN RPAREN
| NEW (INT | IDENTIFIER) (LBRAC RBRAC)?
;
type
: (INT | IDENTIFIER) (LBRAC RBRAC)?
| VOID
;
问题是,在树语法中,您执行以下操作(我相信是3次):
^(扩展标识符)
部分错误:您需要将树括在括号中,然后使其成为可选的:
classdeclaration
: ^(CLASS ... (^(EXTENDS IDENTIFIER))? ... )
;
然而,如果仅此而已,那就太容易了,不是吗?:)
当您修复上面提到的问题时,ANTLR会抱怨在试图从树语法生成树遍历器时树语法是不明确的。ANTLR将向您投掷以下物品:
错误(211):J0_SemanticalAnalysis.g:61:26:[致命]规则分配具有非LL(*)
可从alts 1,2访问的递归规则调用导致的决策。解决
左因子分解或使用语法谓词或使用backtrack=true选项
它抱怨语法中的赋值规则:
assignment
: ^(ASSIGN IDENTIFIER+ expression? expression)
;
由于ANTLR是一个LL解析器生成器1,它从左到右解析令牌。因此,IDENTIFIER+表达式中的可选表达式是什么?表达式
使语法不明确。通过将?
移动到最后一个表达式来解决此问题:
assignment
: ^(ASSIGN IDENTIFIER+ expression expression?)
;
1不要让名称ANTLR中的最后两个字母误导您,它们代表语言识别,而不是它生成的解析器类 如果您也发布整个错误消息,我们将能够提供更多帮助:)不幸的是,Eclipse的ANTLR插件显然没有给出错误消息的行号。我得到的只是一个非常隐晦的:java.lang.NoSuchFieldError:OfficingToken。您是否尝试过删除/注释部分以缩小导致问题的原因?输入时不会生成错误。错误源于antlr试图生成与语法对应的Java文件时。@Bart谢谢。非常感谢!你关于antlr创建TL语言的博客是我在这个项目中的参考资料之一!非常感谢你!现在一切都建立起来了,我可以进入下一阶段了!
assignment
: ^(ASSIGN IDENTIFIER+ expression expression?)
;