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我应该使用 Sablecc 为 MiniPython 编写一个 .grammar 文件。我遇到了这些转变/减少冲突：

shift/reduce conflict in state [stack: TIf PTpower *] on TMult in {
     [ PMltp = * TMult PTopower Mltp ] (shift)
     [ PMlpt = * ] followed by TMult (reduce)
}

shift/reduce conflict in state [stack: TIf PTopower *] on TDiv in {
     [ PMltp = * TDiv PTopower Mltp ] (shift)
     [ PMltp = * ] followed by TDiv (reduce)
}

一些令牌是：

id = letter (letter | digit)*;
digit = ['0' .. '9'];
letter = ['a' .. 'z']|['A' .. 'Z'];
pow = '**';
mult = '*';
div = '/';
plus = '+';
minus = '-';
assert = 'assert';
l_par = '(';
r_par = ')';
l_bra = '[';
r_bra = ']';

我的 .grammar 文件的一部分是这样的：

expression = multiplication exprsn;

exprsn =   {addition} plus multiplication exprsn
         | {subtraction} minus multiplication exprsn
         | {empty};

topower = something tpwr;

tpwr =   {topower} pow something tpwr
       | {empty};

multiplication = topower mltp;

mltp =   {multiplication} mult topower mltp
       | {division} div topower mltp
       | {empty};

something =   {id} id
            | {parexp} l_par expression r_par
            | {fcall} functioncall
            | {value} value
            | {list} id l_bra expression r_bra
            | {other} l_bra value comval* r_bra
            | {assert} assert expression comexpr?;

comexpr = comma expression;

这是我尝试消除左递归后的语法。我注意到如果我assert从生产中删除规则something，我不会遇到任何冲突。此外，{empty}从exprsn,tpwr和mltprules 中删除规则不会给我带来任何冲突，但我认为这不是解决此问题的正确方法。

任何提示将不胜感激。

更新：这是整个语法，在删除左递归之前，根据要求：

Package minipython;

Helpers
    digit = ['0' .. '9'];
    letter = ['a' .. 'z']|['A' .. 'Z']; 
    cr = 13; 
    lf = 10;
    all = [0..127]; 
    eol = lf | cr | cr lf ;
    not_eol = [all - [cr + lf]]; 

Tokens
    tab = 9;
    plus = '+';
    dot = '.';
    pow = '**';
    minus = '-';
    mult = '*';
    div = '/';
    eq = '=';
    minuseq = '-=';
    diveq = '/=';
    exclam = '!';
    def = 'def';
    equal = '==';
    nequal = '!=';
    l_par = '(';
    r_par = ')';
    l_bra = '[';
    r_bra = ']';
    comma= ',';
    qmark = '?';
    gqmark = ';';
    assert = 'assert';
    if = 'if';
    while = 'while';
    for = 'for';
    in = 'in';
    print = 'print';
    return = 'return';
    importkn = 'import';
    as = 'as';
    from = 'from';
    less = '<';
    great = '>';
    true = 'true';
    semi = ':';
    false = 'false';
    quote = '"';
    blank = (' ' | lf | cr);
    line_comment = '#' not_eol* eol; 
    number = digit+ | (digit+ '.' digit+);
    id = letter (letter | digit)*;
    string = '"'not_eol* '"';
    cstring = ''' letter ''';

Ignored Tokens
    blank, line_comment;

Productions
program = commands*;

commands =   {stmt} statement
           | {func} function;

function = def id l_par argument? r_par semi statement;

argument = id eqval? ceidv*;

eqval = eq value;

ceidv = comma id eqval?;

statement =   {if} tab* if comparison semi statement
            | {while} tab* while comparison semi statement
            | {for} tab* for [id1]:id in [id2]:id semi statement
            | {return} tab* return expression
            | {print} tab* print expression comexpr*
            | {assign} tab* id eq expression
            | {minassign} tab* id minuseq expression
            | {divassign} tab* id diveq expression
            | {list} tab* id l_bra [ex1]:expression r_bra eq [ex2]:expression
            | {fcall} tab* functioncall
            | {import} import;

comexpr = comma expression;

expression =   {multiplication} multiplication
             | {addition} expression plus multiplication
             | {subtraction} expression minus multiplication;

topower =   {smth} something
          | {power} topower pow something;

something =   {id} id
            | {parexp} l_par expression r_par
            | {fcall} functioncall
            | {value} value
            | {list} id l_bra expression r_bra
            | {assert} assert expression comexpr?
            | {other} l_bra value comval* r_bra;

comval = comma value;   

multiplication =   {power} topower
                 | {multiplication} multiplication mult topower
                 | {division} multiplication div topower;

import =   {import} importkn module asid? comod*
         | {from} from module importkn id asid? comid*;

asid = as id;

comod = comma module asid?;

comid = comma id asid?;

module = idot* id;

idot = id dot;

comparison =   {true} true
             | {false} false
             | {greater} [ex1]:expression great [ex2]:expression
             | {lesser} [ex1]:expression less [ex2]:expression
             | {equals} [ex1]:expression equal [ex2]:expression
             | {nequals} [ex1]:expression nequal [ex2]:expression;

functioncall = id l_par arglist? r_par;

arglist = expression comexpr*;

value =   {fcall} id dot functioncall
        | {numb} number
        | {str} string
        | {cstr} cstring;

现在的移位/减少冲突是：

shift/reduce conflict in state [stack: TIf PTopower *] on TPow in {
     [ PMultiplication - PTopower * ] followed by TPow (reduce),
     [ PTopower = PTopower * TPow PSomething ] (shift)
}
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1 回答 1

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（注意：这个答案是从原始语法中得出的，而不是从尝试删除左递归中得出的，这有其他问题。没有必要从提供给 LALR(1) 解析器生成器的语法中删除左递归像 SableCC。）

实际上，基本问题是生产：

something = {assert} assert expression comexpr?

这个产生式很奇怪，部分原因是非终结符（“某物”）的名称没有提供任何关于它是什么的暗示，但主要是因为人们通常期望assert expression它是一个语句，而不是表达式的一部分。并且something显然源自expression：

expression = multiplication
multiplication = topower
topower = something

但assert生产以expression. 这会导致歧义，因为

assert 4 + 3

可以解析为：（为简洁起见省略了一些步骤）：

expression = expression     plus    multiplication
                |             |           |
                V             |           |
            something         |           |
                |             |           |
                V             |           |
          assert expression   |           |
             |        |       |           |
             |        V       V           V
          assert      4       +           3

或者，更自然地，如：

expression =          something
                          |
                          V
             assert                expression
                |                      |
                |                      V
                |          expression plus multiplication
                |               |       |        |
                |               V       V        V
             assert             4       +        3

第一次解析似乎不太可能，因为assert（据我猜测）实际上并没有返回一个值。（尽管如果运算符是比较而不是加法，则第二个会更自然。）

在没有看到您要解析的语言的定义的情况下，我无法就如何解决这个问题提供具体建议，但我倾向于assert发表声明，并重命名something为更具描述性的名称（“术语”是常见，尽管我通常使用“原子”）。

于 2018-11-06T18:58:44.643 回答