Yacc 程序无法识别函数声明
我认为我的程序应该能够将以下内容识别为函数声明：
int fn(int i) { int n; return; }
但事实并非如此。
这是我的 yacc 文件的相关部分：
program : declaration_list ;
declaration_list : declaration_list declaration | declaration ;
declaration : var_declaration
| fun_declaration
| '$' { printTable();};
var_declaration : type_specifier ID ';' {$2->value = 0; $2->arraysize = 0;};
| type_specifier ID '[' NUM ']' ';' {$2->arraysize = $4;printf("Array size is %d", $2->arraysize);} ;
type_specifier : INT | VOID ;
fun_declaration : type_specifier ID '(' params ')' compound_stmt {printf("function declaration\n"); printf("Parameters: \n", $2->args); } ;
params : param_list | VOID ;
param_list : param_list ',' param
| param ;
param : type_specifier ID | type_specifier ID '[' ']' ;
compound_stmt : '{' local_declarations statement_list '}' {printf("exiting scope\n"); } ;
local_declarations : local_declarations var_declaration
| /* empty */ ;
statement_list : statement_list statement
| /* empty */ ;
statement : expression_stmt
| compound_stmt
| selection_stmt
| iteration_stmt
| return_stmt ;
expression_stmt : expression ';'
| ';' ;
selection_stmt : IF '(' expression ')' statement
| IF '(' expression ')' statement ELSE statement ;
iteration_stmt : WHILE '(' expression ')' statement ;
return_stmt : RETURN ';' | RETURN expression ';' ;
为什么它无法识别它？
I think my program should be able to recognize the following as a function declaration:
int fn(int i) { int n; return; }
but it doesn't.
Here's the relevant part of my yacc file:
program : declaration_list ;
declaration_list : declaration_list declaration | declaration ;
declaration : var_declaration
| fun_declaration
| '
Why does it not recognize it?
{ printTable();};
var_declaration : type_specifier ID ';' {$2->value = 0; $2->arraysize = 0;};
| type_specifier ID '[' NUM ']' ';' {$2->arraysize = $4;printf("Array size is %d", $2->arraysize);} ;
type_specifier : INT | VOID ;
fun_declaration : type_specifier ID '(' params ')' compound_stmt {printf("function declaration\n"); printf("Parameters: \n", $2->args); } ;
params : param_list | VOID ;
param_list : param_list ',' param
| param ;
param : type_specifier ID | type_specifier ID '[' ']' ;
compound_stmt : '{' local_declarations statement_list '}' {printf("exiting scope\n"); } ;
local_declarations : local_declarations var_declaration
| /* empty */ ;
statement_list : statement_list statement
| /* empty */ ;
statement : expression_stmt
| compound_stmt
| selection_stmt
| iteration_stmt
| return_stmt ;
expression_stmt : expression ';'
| ';' ;
selection_stmt : IF '(' expression ')' statement
| IF '(' expression ')' statement ELSE statement ;
iteration_stmt : WHILE '(' expression ')' statement ;
return_stmt : RETURN ';' | RETURN expression ';' ;
Why does it not recognize it?
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评论（3）
这不是您问题的答案，而是调试建议。
在每个规则之后，添加一个
printf语句，告诉您该规则已匹配。看看我的回答 获取提示。然后通过语法运行您的输入，看看它在做什么。提出问题的另一个提示是尝试将问题简化为重现错误的最简单情况。对于上述语法，在发布之前删除您不希望与此特定示例匹配的所有规则。另外，发布一个完整的程序，并带有
%token声明，以便人们可以编译它并亲自尝试。更好的是也发布词法分析器。This isn't an answer to your question but a suggestion for debugging.
After each of the rules, add a
printfstatement which tells you that the rule has been matched. Take a look at my answer here for hints. Then run your input through the grammar and see what it is doing.Another hint, for asking questions, is to try to reduce your problem to the simplest case which reproduces the error. In the case of the above grammar, strip out all of the rules which you don't expect to match this particular example before posting it. Also, post a complete program, with the
%tokendeclarations, so people can compile it and try it out for themselves. Even better is to post the lexer too.我认为你需要类似的东西：
type_specifier : INT |空白 ;
I think you need something like:
type_specifier : INT | VOID ;
只需查看您所在的行
local_declarations : local_declarations var_declaration | /* empty */ ;该行对我来说看起来很麻烦，因为您的它就像一个递归语句...
希望这能给您提示，
此致，
汤姆.
Just looked at the line where you have
local_declarations : local_declarations var_declaration | /* empty */ ;That line looks troublesome to me, since you have it like a recursive statement...
Hope this gives you the hint,
Best regards,
Tom.