Bison shift/reduce错误
Bison shift/reduce错误
提问于 2015-03-21 23:08:11
我在野牛中有一个错误，我不知道如何修复。
当我试图编译我的语法时，我得到了1个shift/reduce错误：
START: type_specifier MAIN LBR_R RBR_R LBR_C program RBR_C
program: statement ;
type_specifier: INT | DOUBLE | VOID;
var_declaration: type_specifier var_declist SECOLON;
var_declist: var_declist COMMA var_intialise | var_intialise;
var_intialise: var_ID | var_ID EQL simple_expression;
assign : var_ID EQL simple_expression SECOLON;
statement: expression_stmt | local_declaration | selection_stmt | iteration_stmt| return_stmt | break_stmt;
local_declaration: var_declaration | assign;
expression_stmt: expression SECOLON | SECOLON;
selection_stmt: IF LBR_R simple_expression RBR_R LBR_C statement RBR_C;
iteration_stmt: WHILE LBR_R simple_expression RBR_R LBR_C statement RBR_C;
return_stmt: RETURN SECOLON | RETURN expression SECOLON;
break_stmt: BREAK SECOLON;
expression: var_ID EQL expression | var_ID INCR_EQL expression | var_ID DECR_EQL expression | var_ID INCR | var_ID DECR | simple_expression;
simple_expression: simple_expression OR and_expression | and_expression;
and_expression: and_expression AND unary_rel_expression | unary_rel_expression;
unary_rel_expression: NOT rel_expression | rel_expression;
rel_expression: sum_expression relop sum_expression | sum_expression;
relop: LSS_EQL | LSS | GR | GR_EQL | EQUIV | NOT EQL;
sum_expression: sum_expression sumop term | term;
sumop: ADD | SUB;
term: term mulop unary_expression | unary_expression;
mulop: MULT | DIV | RMDR;
unary_expression: factor;
factor: immutable | var_ID;
var_ID: ID;
immutable: LBR_R expression RBR_R | call | constant;
call: ID LBR_R args RBR_R;
args: | arg_list;
arg_list: arg_list COMMA expression | expression;
constant: INT | CHAR | STRING; 我得到1个班次/减少在
状态102冲突:1个shift/reduce
具体状态为：
state 102
11 assign: var_ID EQL simple_expression . SECOLON
32 expression: simple_expression . [SECOLON]
33 simple_expression: simple_expression . OR and_expression
OR shift, and go to state 63
SECOLON shift, and go to state 111
SECOLON [reduce using rule 32 (expression)]有没有人可以解释一下如何修复这个错误，我知道它与优先级有关。
Stack Overflow用户
发布于 2015-03-22 08:12:03
这与优先级无关。
以下是您的语法的相关子集：
statement: expression_stmt | local_declaration
expression_stmt: expression ';'
expression: var_ID EQL expression | simple_expression
local_declaration: assign
assign: var_ID EQL simple_expression ';'这显然是不明确的；var = <simple_expression> ;既可以是expression_stmt，也可以是local_declaration。这种歧义通常会导致reduce/reduce冲突，但在这种情况下，由于处理分号的方式，您会得到shift/reduce冲突。在任何情况下，解决方案都是消除歧义。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/29184258
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