如何在Bison中明确定义一个语句开关分支？

我们希望定义一个分支，其中分支的定义被大括号包围，但是对于单个语句分支，大括号可以省略，比如单个语句

if

块

所需语法的示例（为清晰起见，以白色分隔）：

我有以下语法，但它有移位/减少冲突：

%token <union_string> T_ID   "identifier"

%% // begin rules

program:
    statement_list
    ;

statement_list:
    statement_list statement
    | empty
    ;

statement:
    T_ID //added for completeness
    | "switch" switch_branch_list //the subject of the switch is elided for brevity
    ;

switch_branch_list:
    switch_branch_list "|" switch_branch
    | switch_branch
    ;

switch_branch:
    T_ID statement
    | T_ID "{" statement_list "}"
    ;

empty:
    ;

...
State 9

    5 statement: "switch" switch_branch_list .
    6 switch_branch_list: switch_branch_list . "|" switch_branch

    "|"  shift, and go to state 13

    "|"       [reduce using rule 5 (statement)]
    $default  reduce using rule 5 (statement)
...

---

如何解决给定语法中的shift/reduce冲突？

没有任何东西会终止switch语句，因此如果switch分支的目标是另一个switch语句，则程序将变得不明确

我建议不要让switch（或其他复合语句）成为分支的直接目标。坚持用大括号把它围起来

所以你会得到这样的结果：

statement:          simple_statement | compound_statement
simple_statement:   expr
                    | '{' statement_list '}'
                    | ...
compound_statement: switch_statement
                    | ...
statement_list      /* empty */ | statement_list statement  

switch_statement:   "switch" expr branch_list
branch_list:        branch | branch_list '|' branch
branch:             T_ID simple_statement

注意，我将块语句语法（

'{'statement\u list'}'

）作为

简单语句

的产品之一。通常情况下，这就是它的含义，就像括号中的表达式是标准表达式语法中

术语

的产物一样。但这只是一个建议

statement:          simple_statement | compound_statement
simple_statement:   expr
                    | '{' statement_list '}'
                    | ...
compound_statement: switch_statement
                    | ...
statement_list      /* empty */ | statement_list statement  

switch_statement:   "switch" expr branch_list
branch_list:        branch | branch_list '|' branch
branch:             T_ID simple_statement