如果前瞻标记是给定值,是否有办法指定 Bison 规则不匹配?
我目前有以下野牛语法(简化):
var_decl:
type ident
{
$$ = new NVariableDeclaration(*$1, *$2);
} |
type ident ASSIGN_EQUAL expr
{
$$ = new NVariableDeclaration(*$1, *$2, $4);
} |
type CURVED_OPEN STAR ident CURVED_CLOSE CURVED_OPEN func_decl_args CURVED_CLOSE
{
$$ = new NVariableDeclaration(*(new NFunctionPointerType(*$1, *$7)) /* TODO: free this memory */, *$4);
} |
type CURVED_OPEN STAR ident CURVED_CLOSE CURVED_OPEN func_decl_args CURVED_CLOSE ASSIGN_EQUAL expr
{
$$ = new NVariableDeclaration(*(new NFunctionPointerType(*$1, *$7)) /* TODO: free this memory */, *$4, $10);
} ;
...
deref:
STAR ident
{
$$ = new NDereferenceOperator(*$<ident>2);
} |
...
type:
ident
{
$$ = new NType($<type>1->name, 0, false);
delete $1;
} |
... ;
...
expr:
deref
{
$$ = $1;
} |
...
ident
{
$<ident>$ = $1;
} |
...
ident CURVED_OPEN call_args CURVED_CLOSE
{
$$ = new NMethodCall(*$1, *$3);
delete $3;
} |
...
CURVED_OPEN expr CURVED_CLOSE
{
$$ = $2;
} ;
...
call_args:
/* empty */
{
$$ = new ExpressionList();
} |
expr
{
$$ = new ExpressionList();
$$->push_back($1);
} |
call_args COMMA expr
{
$1->push_back($3);
} ;
问题是解析时:
void (*ident)(char* some_arg);
它看到 void (*ident) 并推断它必须是函数调用而不是函数声明。 有没有办法我可以告诉 Bison,它应该倾向于向前看以匹配 var_decl 而不是将 *ident 和 void 减少为 derefs 和 exprs?