这个语法怎么有歧义?

Question

我正在用Jison编写一个简单的表达式解析器。下面是我的语法:

{
    "operators": [
        ["left", "+", "-"],
        ["left", "*", "/", "%"]
    ],
    "bnf": {
        "program": [
            ["statement EOF", "return $1;"]
        ],
        "statement": [
            ["expression NEWLINE", "$$ = $1 + ';';"]
        ],
        "expression": [
            ["NUMBER",                       "$$ = yytext;"],
            ["expression binary expression", "$$ = $1 + $2 + $3;"]
        ],
        "binary": [
            ["+",              "$$ = ' + ';"],
            ["-",              "$$ = ' - ';"],
            ["*",              "$$ = ' * ';"],
            ["/",              "$$ = ' / ';"],
            ["%",              "$$ = ' % ';"],
            ["binary NEWLINE", "$$ = $1;"]
        ]
    }
}

当我尝试运行它时，它给了我以下错误:

Conflict in grammar: multiple actions possible when lookahead token is + in state
13
- reduce by rule: expression -> expression binary expression
- shift token (then go to state 8)
Conflict in grammar: multiple actions possible when lookahead token is - in state
13
- reduce by rule: expression -> expression binary expression
- shift token (then go to state 9)
Conflict in grammar: multiple actions possible when lookahead token is * in state
13
- reduce by rule: expression -> expression binary expression
- shift token (then go to state 10)
Conflict in grammar: multiple actions possible when lookahead token is / in state
13
- reduce by rule: expression -> expression binary expression
- shift token (then go to state 11)
Conflict in grammar: multiple actions possible when lookahead token is % in state
13
- reduce by rule: expression -> expression binary expression
- shift token (then go to state 12)
States with conflicts:
State 13
  expression -> expression binary expression . #lookaheads= NEWLINE + - * / %
  expression -> expression .binary expression
  binary -> .+
  binary -> .-
  binary -> .*
  binary -> ./
  binary -> .%
  binary -> .binary NEWLINE

然而，它最终仍然产生正确的输出。例如，2 + 3 * 5 / 7 % 11被正确地翻译成2 + 3 * 5 / 7 % 11;。

在我看来，我的语法似乎是明确的，那么Jison为什么要抱怨呢?

Update:正如@icktoofay解释的，这是一个运算符结合性问题。通过将操作符解析为非终端符号，操作符优先级和结合性信息将丢失。因此，我解决了这个问题，如下:

{
    "operators": [
        ["left", "+", "-"],
        ["left", "*", "/", "%"]
    ],
    "bnf": {
        "program": [
            ["statement EOF", "return $1;"]
        ],
        "statement": [
            ["expression NEWLINE", "$$ = $1 + ';';"]
        ],
        "expression": [
            ["NUMBER",                          "$$ = yytext;"],
            ["expression + expression",         "$$ = $1 + ' + ' + $3;"],
            ["expression - expression",         "$$ = $1 + ' - ' + $3;"],
            ["expression * expression",         "$$ = $1 + ' * ' + $3;"],
            ["expression / expression",         "$$ = $1 + ' / ' + $3;"],
            ["expression % expression",         "$$ = $1 + ' % ' + $3;"],
            ["expression + NEWLINE expression", "$$ = $1 + ' + ' + $4;"],
            ["expression - NEWLINE expression", "$$ = $1 + ' - ' + $4;"],
            ["expression * NEWLINE expression", "$$ = $1 + ' * ' + $4;"],
            ["expression / NEWLINE expression", "$$ = $1 + ' / ' + $4;"],
            ["expression % NEWLINE expression", "$$ = $1 + ' % ' + $4;"]
        ]
    }
}

也就是说，该语法只允许一个可选的换行符跟在二进制操作符后面。如何重写它以允许任意数量的换行符跟随二进制操作符?也必须有一些方法，我不必为每个操作符写两个规则。

Answer 1

我不是很熟悉Jison，但是看起来您定义的规则是这样的:

expression ::= number;
expression ::= expression binary expression;

考虑表达式1 - 2 - 3。这可以解释为(1 - 2) - 3或1 - (2 - 3)。是哪一个?你的语法含糊不清。一般的数学规则说它应该是左结合律。你需要让你的语法反映出:

expression ::= number;
expression ::= expression binary number;