问题
I am using CUP to generate a parser, and I want an empty file to be an acceptable program. I have tried add the empty case to my start symbol, based off the response to a similar question here.
start with prog;
/* The grammar rules */
prog ::= class_block:cb
| class_block:cb stmts:sb
| stmts:sb
| // desired empty case
;
Including the desired empty case gives me the following error:
parser.java:516: error: incompatible types: Object cannot be converted to Symbol
CUP$parser$result = parser.getSymbolFactory().newSymbol("prog",0, ((java_cup.runtime.Symbol)CUP$parser$stack.peek()), RESULT);
How can I modify my grammar so that the parser accepts an empty file? I am using Jflex as my lexer, and ComplexSymbolFactory as the type of the symbols.
EDIT: I've confirmed that the grammar above is the correct way to include empty. However, ComplexSymbolFactory is having problems converting the empty object to a symbol. I get this error even when running this example from the official CUP website.
回答1:
I downloaded the .jars from a friend github project and problem solved.
You can see it's a BUG here: https://github.com/jflex-de/jflex/issues/384
Try to downgrade the version.
回答2:
I don't know about any bugs, but I know you can rewrite your grammar to make the empty case work. for example:
Prog ::=
class_block:cb statement:s
| /* the empty production */
;
I have tried this, and it works for empty productions. Having a production where it is a block and then a statement, or a statement, or an empty, conflicts. there isn't a way to tell during the generation for the parser at the program grammar rules which is a statement and what is an empty. Changing it only being a block and a statement, or an empty clarifies that.
来源:https://stackoverflow.com/questions/52862424/how-to-match-the-empty-case-in-cup-parser-grammar