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我在我的语法上运行 Python 的 PLY 库。它似乎可以编译，并且库不会提醒我任何 shift/reduce 或 reduce/reduce 错误。但是，在一个非常简单的示例上运行会引发错误。

挖掘parser.out文件显示错误发生在程序的最后：

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

其中状态 113 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，解析器应该减少到fn_def.

$end为什么到达令牌时不会发生reduce ？

（如果有帮助，我可以粘贴我的语法，虽然它可能有点长。）

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>
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1 回答 1

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似乎它需要一个分号，并且它有输入结束。不看语法就很难说更多。

语法的相关部分：

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

这些是唯一出现在右侧的fn_def作品。definition

显然definition只能归结为statement前瞻记号为 时SEMICOLON。由于fn_def只能出现在有效程序中可以立即归约到的地方definition（唯一的产生是单位产生），fn_def所以后面必须跟一个SEMICOLON. 所以你的解析器是正确的，你的样本输入是不合语法的。

fn_def只有一个产品：

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

其中很明显，最后一项fn_def是 a statement。某些语句 ( definitionand expression) 必须以;;结尾。如果fn_def'sstatement是其中之一（可能是一个表达式，因为它看起来不像单个定义会产生有趣的函数体），那么fn_def将必须用两个分号编写。我怀疑这是否是你想要的。

definition以 a statement（如果它是 a fn_def）或以 an expression（如果它是 a var_def）结尾。您已尝试定义statement它以使其具有自定界性（即，如果它不以}终止 a结尾，则以分号结尾compound_statement。因此，fn_def已经以分号或右大括号结尾，并且不需要另一个分号. var_def, 另一方面，以表达式结尾，因此也是如此。因此，一种解决方案是将结束的分号推入var_def.

编辑评论，与所问的具体问题无关：

事实上，除了你自己的审美之外，没有明显的理由需要将循环体或条件体限制为复合语句。如果您允许 lambda 主体成为非复合语句，则没有明显的理由限制 for 循环。可以使语法以任何一种方式工作。

于 2013-07-22T14:23:33.320 回答