e2中let e1的明确gramar

Question

我试图创造一种让语法我的想法是这样的

start : let
let : "let" ID ("=" let)? in let | atom
atom : ANYTHING | "(" let ")"
ID : /[a-z]+/

其思想是解析这样的表达式let A = B in C或let A in B或两者混合let f = let x in x + 1 in f(1)。我还想支持括号来消除歧义，比如let A = (let b in b + 1) in A(1) + 1

我使用的是lark，带有LALR解析器，但我一直在努力学习语法，无法为此定义明确的语法

我试过了

from lark import Lark, Transformer as LarkTransformer


grammar = """
    start : expr
    expr : LET ID (EQUAL exprcont)? IN exprcont | exprcont
    exprcont : ANYTHING | LPAR expr RPAR | expr
    ANYTHING.0 : /.+/
    LET : "let"
    IN : "in"
    ID : /[a-z_][a-z0-9_]*/
    EQUAL : "="
    LPAR.10 : "("
    RPAR.10 : ")"

    %import common.WS
    %ignore WS
"""

let_parser = Lark(grammar, parser="lalr")

print(let_parser.parse("let a = 1 in let b = 2 in a + b").pretty())

但是我有很多减少错误的方法

Traceback (most recent call last):
  File "/Users/gecko/code/lampycode/letparser.py", line 55, in <module>
    let_parser = Lark(grammar, parser="lalr")
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/lark.py", line 339, in __init__
    self.parser = self._build_parser()
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/lark.py", line 373, in _build_parser
    return self.parser_class(self.lexer_conf, parser_conf, options=self.options)
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/parser_frontends.py", line 145, in __init__
    self.parser = LALR_Parser(parser_conf, debug=debug)
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/parsers/lalr_parser.py", line 17, in __init__
    analysis.compute_lalr()
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/parsers/lalr_analysis.py", line 304, in compute_lalr
    self.compute_lalr1_states()
  File "/Users/gecko/.pyenv/versions/lampy/lib/python3.9/site-packages/lark/parsers/lalr_analysis.py", line 279, in compute_lalr1_states
    raise GrammarError('\n\n'.join(msgs))
lark.exceptions.GrammarError: Reduce/Reduce collision in Terminal('$END') between the following rules: 
    - <exprcont : expr>
    - <start : expr>

Reduce/Reduce collision in Terminal('IN') between the following rules: 
    - <expr : exprcont>
    - <expr : LET ID IN exprcont>

Reduce/Reduce collision in Terminal('RPAR') between the following rules: 
    - <expr : exprcont>
    - <expr : LET ID IN exprcont>

Reduce/Reduce collision in Terminal('$END') between the following rules: 
    - <expr : exprcont>
    - <expr : LET ID IN exprcont>

Reduce/Reduce collision in Terminal('IN') between the following rules: 
    - <expr : exprcont>
    - <expr : LET ID EQUAL exprcont IN exprcont>

Reduce/Reduce collision in Terminal('RPAR') between the following rules: 
    - <expr : exprcont>
    - <expr : LET ID EQUAL exprcont IN exprcont>

Reduce/Reduce collision in Terminal('$END') between the following rules: 
    - <expr : exprcont>

我不知道如何定义这个语法，这个想法很简单let : "let" ID ("=" let)? "in" let | atom有什么想法吗

Answer 1

如果您想使用像ANYTHING这样的终端，请不要使用lalr。使用earley。（即使如此，它仍然会产生问题）

---

但这并不是造成这些错误的真正原因。问题是expr和exprcont中的相互递归。您只需删除exprcont中的expr：

exprcont : ANYTHING | LPAR expr RPAR

---

但这仍然行不通。（除非您使用parser='earley'和lexer='dynamic_complete'。但这将非常缓慢）。您必须重新设计语法，使其不包含ANYTHING终端

Answer 2

我想问题是

start : expr
    expr : ... | exprcont
    exprcont : ... | expr

这个循环意味着你的语法模棱两可

你能摆脱这个循环吗