解析器规则不会产生无效输入的错误
我发现了一种奇怪的解析器行为，无法向我解释。如果我做错了什么，也许你们中的人可以给我一个提示。
我有关注的语法文件：
grammar test;
/*
* Parser Rules
*/
unary_expr
: unary_operator IDENTIFIER;
unary_operator
: PLUS PLUS;
/*
* Lexer Rules
*/
IDENTIFIER: [a-zA-Z]+;
PLUS: '+';
NEWLINE
: ( '\r'? '\n' | '\r' )+ -> skip;
测试此语法时，
++abc++def
当我用Lexer
Tokens:
[@0,0:0='+',<2>,1:0]
[@1,1:1='+',<2>,1:1]
[@2,2:4='abc',<1>,1:2]
[@3,5:5='+',<2>,1:5]
[@4,6:6='+',<2>,1:6]
[@5,7:9='def',<1>,1:7]
[@6,12:11='<EOF>',<-1>,2:0]
将正确检测到所有令牌，但是解析树绝对不能覆盖所有检测到的令牌！
Parse Tree:
unary_expr (
unary_operator (
"+"
"+"
)
"abc"
)
我在做什么错？
I have discovered a strange parser behavior and cannot explain it to me. Maybe someone of you can give me a hint if I am doing something wrong.
I have following grammer file:
grammar test;
/*
* Parser Rules
*/
unary_expr
: unary_operator IDENTIFIER;
unary_operator
: PLUS PLUS;
/*
* Lexer Rules
*/
IDENTIFIER: [a-zA-Z]+;
PLUS: '+';
NEWLINE
: ( '\r'? '\n' | '\r' )+ -> skip;
When I test this grammar with
++abc++def
the lexer will detect all tokens properly
Tokens:
[@0,0:0='+',<2>,1:0]
[@1,1:1='+',<2>,1:1]
[@2,2:4='abc',<1>,1:2]
[@3,5:5='+',<2>,1:5]
[@4,6:6='+',<2>,1:6]
[@5,7:9='def',<1>,1:7]
[@6,12:11='<EOF>',<-1>,2:0]
But the parse tree definitely does not cover all detected tokens!
Parse Tree:
unary_expr (
unary_operator (
"+"
"+"
)
"abc"
)
What am I doing wrong?
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如果您告诉Antlr匹配Parser Rule
unary_expr消耗的任何代币，那么这正是Antlr所做的。如果要强制ANTLR消耗 all 令牌，请在解析器规则结束时添加
eof令牌：现在您会看到您的预期错误：
If you tell ANTLR to match whatever tokens are consumed by parser rule
unary_expr, then that is exactly what ANTLR does.If you want to force ANTLR to consume all tokens, add the
EOFtoken at the end of your parser rule:Now you will see your expected error: