我想知道如何评估 ifdef ... else ... end 语句,它可以放在代码的任何地方。我从一个简单的例子开始,它实现了 2 个基本函数 add(p1,p2) 和 diff(p1,p2),其中 p1 和 p2 是字符串。它只在 p1 和 p2 之间添加一个 + 或一个 -。这是我的语法:
grammar ifdef;
options {
language = Java;
output = AST;
ASTLabelType=CommonTree;
}
tokens
{
EQUAL = '=' ;
HASH = '#' ;
DBLEQUOTE = '"' ;
SEMICOLON = ';' ;
}
@header {
package Grammar;
import java.util.Map;
import java.util.HashMap;
}
@lexer::header {
package Grammar;
}
@members {
private Map<String, String> strMapID = new HashMap<String, String>();
private Map<String, String> strMapDefine = new HashMap<String, String>();
}
rule returns [String strEval]
: { StringBuilder strBuilder = new StringBuilder(); }
( command
{ if ( $command.str != "" ) {
strBuilder.append( $command.str );
strBuilder.append( "\n" );
}
}
)+ EOF
{ $strEval = strBuilder.toString(); }
;
command returns [String str]
: define { $str=""; }
| undef { $str=""; }
| set { $str=""; }
| function { $str = $function.str; }
| ifdef { $str=""; }
;
define
: HASH 'define' ID
{ strMapDefine.put($ID.text, $ID.text); } // save define ID into hash table
;
undef
: HASH 'undef' ID
{ if ( strMapDefine.containsKey($ID.text) ) {
strMapDefine.remove($ID.text); // undef ID in hash table
}
}
;
set
: 'set' ID EQUAL string SEMICOLON
{ strMapID.put($ID.text, $string.text); } // save ID,string definition into hash table
;
string
: DBLEQUOTE expr DBLEQUOTE
;
function returns [String str]
: add { $str = $add.str; }
| diff { $str = $diff.str; }
;
add returns [String str]
: 'add' '(' p1=param ',' p2=param ')'
{ StringBuilder strBuilder = new StringBuilder();
strBuilder.append( $p1.str );
strBuilder.append( "+" );
strBuilder.append( $p2.str );
$str = strBuilder.toString();
}
;
diff returns [String str]
: 'diff' '(' p1=param ',' p2=param ')'
{ StringBuilder strBuilder = new StringBuilder();
strBuilder.append( $p1.str );
strBuilder.append( "-" );
strBuilder.append( $p2.str );
$str = strBuilder.toString();
}
;
param returns [String str]
: ID { $str = strMapID.get($ID.text); } // assign ID definition to str
| string { $str = $string.text; }
| function { $str = $function.str; }
;
ifdef
: { Boolean bElse= false;
StringBuilder strBuilder = new StringBuilder(); }
HASH 'ifdef' '(' ID ')' c1=expr ( HASH 'else' c2=expr { bElse= true; } )? HASH 'end'
{ if ( strMapDefine.containsKey($ID.text) ) {
strBuilder.append($c1.text);
}
else {
if ( bElse ) {
strBuilder.append($c2.text);
}
}
System.out.println("ifdef content is : " + strBuilder.toString() + "\n" );
}
;
expr
: .+
;
ID
: ('a'..'z' | 'A'..'Z')+
;
WS
: ( ' ' | '\t' | '\n' | '\r' ) {$channel=HIDDEN;}
;
和java主类
package Grammar;
import java.io.IOException;
import org.antlr.runtime.ANTLRFileStream;
import org.antlr.runtime.CharStream;
import org.antlr.runtime.CommonTokenStream;
import org.antlr.runtime.RecognitionException;
import org.antlr.runtime.TokenStream;
import Grammar.ifdefParser.rule_return;
public class mainifdef {
public static void main(String[] args) throws RecognitionException {
CharStream stream=null;
try {
stream = new ANTLRFileStream("src/input/test.txt");
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
ifdefLexer lexer = new ifdefLexer(stream);
TokenStream tokenStream = new CommonTokenStream(lexer);
ifdefParser parser = new ifdefParser(tokenStream);
rule_return evaluator = parser.rule();
System.out.println("Parsing Tree is \n" + evaluator.tree.toStringTree() + "\n");
System.out.println("Evaluation is \n" + evaluator.strEval + "\n");
}
}
输入文件内容如下:
#undef Z
set AB = "toto";
设置 CD = "titi";差异(CD,“essai”)
#ifdef(Z)添加(#end添加(AB,差异(CD,“essai”))
结果控制台符合预期
ifdef 内容是:
解析树是
#undef Z set AB = "toto" ; 设置 CD = "titi" ; diff (CD, "essai") #ifdef (Z) add (#end add (AB, diff(CD, "essai"))评价为
“titi”-“essai”
“toto”+“titi”-“essai”
我的问题是如何评估应该给出相同结果的输入文本:
#undef Z
set AB = "toto";
设置 CD = "titi";差异(CD,“essai”)
#ifdef(Z)添加(#end添加(AB,#ifdef(Z)添加(#else diff(#end CD,“essai”))
结果控制台是(这是正常的 ANTLR 行为):
ifdef 内容是:
src/input/test.txt 行 8:11 在输入 '#' 处没有可行的替代方案
src/input/test.txt 行 8:43 在 'CD'
ifdef content is : diff(解析树是
#undef Z set AB = "toto" ; 设置 CD = "titi" ; diff ( CD , " essai " ) # ifdef ( Z ) add ( # end add ( AB , < unexpected: [@59,94:94='#',< 6 >,8:11], resync=# > <缺少')'> # ifdef ( Z ) add ( # else diff ( # end <缺少EOF >评价为
“titi”-“essai”
“toto”+null
一些从哪里开始的指导将更加感激
问候 JPM