为什么Python在$end令牌上停止解析？

Question

我正在我的语法上运行Python的PLY库。它似乎编译得很好，而且库没有提醒我注意任何移位/减少或减少/减少错误。但是，在一个非常简单的示例上运行会引发一个错误。

深入了解parser.out文件可以发现，错误发生在程序的末尾：

State  : 113
Stack  : DEFN ID ARROW type invariants statement . $end
ERROR: Error  : DEFN ID ARROW type invariants statement . $end
Syntax error at: None

113国是：

state 113

    (16) fn_def -> DEFN ID ARROW type invariants statement .
    (24) tilde_loop -> statement . TILDE constant TILDE

    SEMICOLON       reduce using rule 16 (fn_def -> DEFN ID ARROW type inva\
riants statement .)
    TILDE           shift and go to state 62

据我所知，解析器应该简化为fn_def。

为什么在到达$end令牌时不会发生减少呢？

(如果语法有帮助的话，我可以粘贴它，尽管它可能有点长。)

语法

Rule 0     S' -> program
Rule 1     program -> statement_list
Rule 2     statement -> definition SEMICOLON
Rule 3     statement -> expression SEMICOLON
Rule 4     statement -> control_statement
Rule 5     statement -> compound_statement
Rule 6     statement -> comment
Rule 7     statement -> empty SEMICOLON
Rule 8     statement_list -> statement_list statement
Rule 9     statement_list -> statement
Rule 10    control_statement -> loop
Rule 11    control_statement -> if_statement
Rule 12    compound_statement -> CURLY_OPEN statement_list CURLY_CLOSE
Rule 13    compound_statement -> CURLY_OPEN CURLY_CLOSE
Rule 14    definition -> fn_def
Rule 15    definition -> var_def
Rule 16    fn_def -> DEFN ID ARROW type invariants statement
Rule 17    var_def -> type assignment
Rule 18    assignment -> ID ASSIGN expression
Rule 19    loop -> for_loop
Rule 20    loop -> foreach_loop
Rule 21    loop -> tilde_loop
Rule 22    for_loop -> FOR statement statement statement compound_statement
Rule 23    foreach_loop -> FOREACH ID IN expression compound_statement
Rule 24    tilde_loop -> statement TILDE constant TILDE
Rule 25    if_statement -> IF condition compound_statement
Rule 26    if_statement -> IF condition compound_statement elseif_list
Rule 27    if_statement -> IF condition compound_statement elseif_list else
Rule 28    if_statement -> ternary
Rule 29    elseif_list -> ELSEIF condition compound_statement
Rule 30    elseif_list -> ELSEIF condition compound_statement elseif_list
Rule 31    else -> ELSE compound_statement
Rule 32    ternary -> condition QUESTION_MARK expression COLON expression
Rule 33    invariants -> invariants invariant
Rule 34    invariants -> invariant
Rule 35    invariants -> NONE
Rule 36    invariant -> type ID COLON invariant_tests
Rule 37    invariant_tests -> invariant_tests COMMA test
Rule 38    invariant_tests -> test
Rule 39    test -> ID operator constant
Rule 40    test -> STAR
Rule 41    expression -> declarator
Rule 42    expression -> assignment
Rule 43    expression -> container
Rule 44    expression -> test
Rule 45    expression -> constant
Rule 46    type -> INT_T
Rule 47    type -> DBL_T
Rule 48    type -> STR_T
Rule 49    type -> list_type
Rule 50    operator -> GT_OP
Rule 51    operator -> LT_OP
Rule 52    operator -> GTE_OP
Rule 53    operator -> LTE_OP
Rule 54    operator -> EQ_OP
Rule 55    operator -> NEQ_OP
Rule 56    constant -> INT
Rule 57    constant -> DBL
Rule 58    constant -> STR
Rule 59    declarator -> ID L_PAREN fn_args R_PAREN
Rule 60    declarator -> ID
Rule 61    fn_args -> fn_args COMMA expression
Rule 62    fn_args -> expression
Rule 63    container -> list
Rule 64    list -> L_BRACE container_items R_BRACE
Rule 65    list_type -> L_BRACE type R_BRACE
Rule 66    container_items -> container_items COMMA container_item
Rule 67    container_items -> container_item
Rule 68    container_item -> expression
Rule 69    container_item -> empty
Rule 70    bool -> TRUE
Rule 71    bool -> FALSE
Rule 72    condition -> bool
Rule 73    comment -> single_comment
Rule 74    single_comment -> COMMENT_LINE
Rule 75    empty -> <empty>

Answer 1

好像它需要一个分号，而且它有输入结束。很难不看语法就说得更多。

你语法的相关部分：

 (2) statement -> definition SEMICOLON
(14) definition -> fn_def

这些是fn_def和definition出现在右侧的唯一产品。

显然，只有当查找令牌是definition时，才能将statement还原为SEMICOLON。因为fn_def只能出现在一个有效的程序中，在那里它可以立即还原为definition (唯一的生产是单位生产)，所以fn_def必须后面跟着一个SEMICOLON。所以你的解析器是正确的，你的样本输入是不语法的。

fn_def只有一个产品：

(16) fn_def -> DEFN ID ARROW type invariants statement

显然，fn_def中的最后一项是statement。有些语句(definition和expression)必须用;来结束；如果fn_def的statement是其中之一(大概是一个表达式，因为它看起来不像是一个有趣的函数体)，那么fn_def必须用两个分号来写。我怀疑这是不是你想要的。

definition以statement (如果是fn_def)或expression (如果是var_def)结尾。您已经尝试过定义statement，以便它是自定界的(也就是说，如果它没有以终止compound_statement的}结尾，则以分号结尾。因此，fn_def已经以分号或结束大括号结尾，不应该需要另一个分号。另一方面，var_def以表达式结尾，因此确实如此。因此，一种解决方案是将结束分号推入var_def。

与被问及的具体问题无关的社论评论：

事实上，除了自己的美学之外，没有明显的理由需要限制循环或条件体来复合语句；如果允许lambda body是非复合语句，则没有明显的理由限制for循环。无论哪种方式，语法都可以发挥作用。