[英]How to resolve shift reduce conflicts in my grammar?
I'm writing a compiler from (reduced) Pascal into ARM asm. 我正在将Pascal(精简版)的编译器编写为ARM asm。 I'm at the second step of the process - after writing lexical analyzer now I'm working on syntax analysis with java cup .
我处于过程的第二步-编写词法分析器之后,现在我正在使用Java cup进行语法分析。
I have written my grammar, but got 5 S/R conflicts, which are all very similar. 我已经编写了语法,但是遇到了5个S / R冲突,它们非常相似。 Example:
例:
Warning : *** Shift/Reduce conflict found in state #150
between assign_stmt ::= val_expr ASSIGN val_expr (*)
and val_expr ::= val_expr (*) LBRACKET val_expr RBRACKET
under symbol LBRACKET
Resolved in favor of shifting
My grammar for this section: 我本节的语法:
assign_stmt ::=
val_expr ASSIGN val_expr;
val_expr ::=
NIL | BOOL_CONST | INT_CONST | CHAR_CONST | PTR val_expr %prec MEM | ADD val_expr %prec UADD |
SUB val_expr %prec USUB | NOT val_expr | val_expr PTR %prec VAL | val_expr MUL val_expr |
val_expr DIV val_expr | val_expr ADD val_expr | val_expr SUB val_expr | val_expr EQU val_expr |
val_expr NEQ val_expr | val_expr LTH val_expr | val_expr GTH val_expr | val_expr LEQ val_expr |
val_expr GEQ val_expr | val_expr AND val_expr | val_expr OR val_expr | IDENTIFIER |
val_expr LBRACKET val_expr RBRACKET | val_expr DOT IDENTIFIER | IDENTIFIER LPARENTHESIS params_list RPARENTHESIS |
LBRACKET type_desc RBRACKET | LPARENTHESIS val_expr RPARENTHESIS
;
How could I eliminate this conflict? 我如何消除这种冲突?
Thanks. 谢谢。
Your grammar is ambiguous, and both right- and left-recursive. 您的语法是模棱两可的,并且都是右递归和左递归的。 From my (limited) knowledge about parsers, I know this is impossible to parse by most parser generators.
根据我对解析器的有限了解,我知道大多数解析器生成器都无法解析。
It's ambiguous because val_expr ADD val_expr SUB val_expr
can be parsed as: 这是模棱两可的,因为
val_expr ADD val_expr SUB val_expr
可以解析为:
ADD
/ \
val_expr SUB
/ \
val_expr val_expr
and 和
SUB
/ \
ADD val_expr
/ \
val_expr val_expr
I've never used Java CUP, but here's how I did a similar thing using another parser generator: 我从未使用过Java CUP,但是这是我使用另一个解析器生成器执行类似操作的方式:
val_expr ::=
expr1 (SUB | ADD | <add all your operators here>) val_expr
| expr1 ;
expr1 ::=
NIL | BOOL_CONST | INT_CONST | CHAR_CONST | <etc> ;
This makes the grammar unambiguous and only right-recursive, which can be handled by all parser generators I know of. 这使得语法是明确的,并且仅是右递归的,可以由我所知道的所有解析器生成器来处理。
A negative aspect of this grammar is that you don't have any precedence, but Java CUP probably has another way to specify precedence. 该语法的不利方面是您没有任何优先级,但是Java CUP可能有另一种指定优先级的方式。
Edit : Fixed first grammar rule. 编辑 :修复了第一个语法规则。
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