shift/reduce 语法与可选分号冲突

Question

我想做这样的事情：

// match used as a statement, semicolon is optional
match (1) {}

// match used as an expression, semicolon is required
1 + match (2) {};

statement
    : expression_without_block T_SEMICOLON
    | expression_with_block
;

expression
    : expression_without_block
    | expression_with_block
;

expression_without_block
    : scalar
    | expression_without_block T_PLUS expression
    | T_PLUS expression
;

expression_with_block
    : T_MATCH T_LEFT expression T_RIGHT T_CURLY_LEFT T_CURLY_RIGHT
;

这个想法是 expression_with_block 不能用在语句的开头，使以下内容明确无误：

match (1) {}
+2;

// Can't mean
(match (1) {} + 2);
// because of the line "expression_without_block T_PLUS expression"

语法导致 shift/reduce 冲突，但我不知道为什么。输出如下（但我不确定如何处理）：

State 11 conflicts: 1 shift/reduce

...

State 11

    5 expression: expression_without_block .
    8 expression_without_block: expression_without_block . T_PLUS expression

    T_PLUS  shift, and go to state 14

    T_PLUS    [reduce using rule 5 (expression)]
    $default  reduce using rule 5 (expression)

The full output file can be found here.

我什至不确定这是正确的方法，因为这样的方法是行不通的：

// `match` can't be at the lhs of an expression, even in sub-expressions
func_call(match (10) {} + 20);

有没有办法在 bison 中实现我在这里描述的内容？我不是野牛专家，所以我真的很感激一些帮助。谢谢！

Answer 1

这是左递归规则发生的一种非常经典的冲突，但由于 shift/reduce 冲突的默认解决方案是进行移位，那么这里一切都会好起来的。

阅读野牛踪迹的方法如下：

T_PLUS  shift, and go to state 14              -- when next token is T_PLUS shift action is selected

T_PLUS    [reduce using rule 5 (expression)]   -- the action between [] was disabled by Bison default conflict resolution

要覆盖 Bison 的默认冲突解决方案（此处不需要），我们可以使用运算符优先级（参见 https://www.gnu.org/software/bison/manual/bison.html#Precedence-Decl) or precedence for non operator (see https://www.gnu.org/software/bison/manual/bison.html#Non-Operators）

shift/reduce 语法与可选分号冲突

shift/reduce conflict in grammar with optional semicolon

yacc

bison