Lark解析器在if/else语句之后计算print语句时引发错误

Question

因此，我正在使用python和lark库制作一种用于解析的编程语言。当我分析以下内容时

if 5 == 4 {
    print("TRUE");
}
else {
    print("FALSE");
}
print("Done!");

它会引发以下错误

PS E:\ParserAndLexer> & C:/Python38/python.exe e:/ParserAndLexer/lite/lite_transformer.py
Traceback (most recent call last):
  File "C:\Python38\lib\site-packages\lark\lexer.py", line 416, in lex
    for x in l.lex(stream, self.root_lexer.newline_types, self.root_lexer.ignore_types):
  File "C:\Python38\lib\site-packages\lark\lexer.py", line 200, in lex
    raise UnexpectedCharacters(stream, line_ctr.char_pos, line_ctr.line, line_ctr.column, allowed=allowed, state=self.state, token_history=last_token and [last_token])
lark.exceptions.UnexpectedCharacters: No terminal defined for 'p' at line 7 col 1

print("HI");
^

Expecting: {'IF'}

Previous tokens: Token('RBRACE', '}')


During handling of the above exception, another exception occurred:

Traceback (most recent call last):
  File "e:/ParserAndLexer/lite/lite_transformer.py", line 80, in <module>
    tree = parser.parse(lite_code)
  File "C:\Python38\lib\site-packages\lark\lark.py", line 464, in parse
    return self.parser.parse(text, start=start)
  File "C:\Python38\lib\site-packages\lark\parser_frontends.py", line 148, in parse
    return self._parse(token_stream, start, set_parser_state)
  File "C:\Python38\lib\site-packages\lark\parser_frontends.py", line 63, in _parse
    return self.parser.parse(input, start, *args)
  File "C:\Python38\lib\site-packages\lark\parsers\lalr_parser.py", line 35, in parse
    return self.parser.parse(*args)
  File "C:\Python38\lib\site-packages\lark\parsers\lalr_parser.py", line 86, in parse
    for token in stream:
  File "C:\Python38\lib\site-packages\lark\indenter.py", line 32, in _process
    for token in stream:
  File "C:\Python38\lib\site-packages\lark\lexer.py", line 431, in lex
    raise UnexpectedToken(t, e.allowed, state=e.state)
lark.exceptions.UnexpectedToken: Unexpected token Token('NAME', 'print') at line 7, column 1.
Expected one of:
        * IF

我不明白为什么会发生这种情况，我的代码是这样的：

from lark import Lark, Transformer, v_args
from lark.indenter import Indenter

class MainIndenter(Indenter):
    NL_type = '_NL'
    OPEN_PAREN_types = ['LPAR', 'LBRACE']
    CLOSE_PAREN_types = ['RPAR', 'RBRACE']
    INDENT_TYPE = '_INDENT'
    DEDENT_type = '_DEDENT'
    tab_len = 8

@v_args(inline=True)
class MainTransformer(Transformer):
    def __init__(self):
        ...

    def number(self, value):
        return Integer(value)

    def string(self, value):
        value = str(value).strip('"')
        return String(value)

    def div(self, val1, val2):
        return Div(val1, val2)

    def print_statement(self, value):
        return Print(value)

    def if_statement(self, expr1, expr2, eval_expr):
        return If(expr1, expr2, eval_expr)
    
    def if_else_statement(self, expr1, expr2, eval_expr, else_statement):
        return If(expr1, expr2, eval_expr, else_statement)
    
    def if_statements(self, *values):
        for value in values:
            value.eval()

    def statement(self, *values):
        for value in values:
            value.eval()

grammar = '''
?start: expr*
      | statement* -> statement
      | if* -> if_statements

?if : "if" expr "==" expr "{" statement+ "}" -> if_statement
    | "if" expr "==" expr "{" expr+ "}" -> if_statement
    | "if" expr "==" expr "{" statement+ "}" "else" "{" statement+ "}" -> if_else_statement

?statement: "print" "(" expr ")" ";"  -> print_statement
          | "input" "(" expr ")" ";"  -> input_statement
          | NAME "=" expr ";"      -> assign_var
          | NAME "=" "input" "(" expr ")" ";" -> var_input_statement

?expr: STRING            -> string
     | NUMBER            -> number
     | NAME              -> get_var
%import common.ESCAPED_STRING -> STRING 
%import common.NUMBER
%import common.CNAME -> NAME
%declare _INDENT _DEDENT
%import common.WS_INLINE
%ignore WS_INLINE
%import common.NEWLINE -> _NL
%ignore _NL
'''
class Print():
    def __init__(self, value):
        self.value = value

    def eval(self):
        return print(self.value.eval())

class Input():
    def __init__(self, value):
        self.value = value

    def eval(self):
        return input(self.value.eval())

class String():
    def __init__(self, value):
        self.value = str(value).strip('"')

    def eval(self):
        return self.value

class Integer():
    def __init__(self, value):
        self.value = int(value)

    def eval(self):
        return self.value

class If():
    def __init__(self, expr1, expr2, eval_expr, else_statement=None):
        self.expr1 = expr1
        self.expr2 = expr2
        self.eval_expr = eval_expr
        self.else_statement = else_statement
        
    def eval(self):
        if self.expr1.eval() == self.expr2.eval():
            return self.eval_expr.eval()
        else:
            if self.else_statement == None:
                return
            else:
                return self.else_statement.eval()

parser = Lark(grammar, parser='lalr', postlex=MainIndenter())
test_input = '''
if 5 == 5 {
    print("True");
}
else {
    print("False");
}
print("Done");
'''

if __name__ == '__main__':
    tree = parser.parse(test_input)
    print(MainTransformer().transform(tree))

Answer 1

我不熟悉lark，但这看起来不对：

?start: expr*
      | statement* -> statement
      | if* -> if_statements

这意味着“将开始规则扩展为零或多个expr、零或多个statements、或零或多个if。这意味着您的语法不支持在试图解析的源字符串中混合这三种产品。如果以if开始，则程序的其余部分必须是全部ifs，因此在{}中抛出一个{}是被禁止的（错误消息说它期待另一个{}）

您可以通过以下方式解决此问题：

?start: stmt*
?stmt: expr
     | statement -> statement
     | if -> if_statements

该语法表示“将开始规则扩展为零或多个stmt，其中stmt被定义为expr、statement或if。通过这种方式，您可以混合和匹配这三种类型的产品

撇开尴尬的命名选择不谈，经过这一短期修复后，语法还有其他明显的缺陷，比如无法支持嵌套的if块。由于您的最终目标不明确，我将避免假设，并保留您当前问题的范围