On Wed, Jan 15, 2014 at 12:41:34PM +0100, Pablo Neira Ayuso wrote:
> On Wed, Jan 15, 2014 at 11:27:17AM +0000, Patrick McHardy wrote:
> > On Wed, Jan 15, 2014 at 11:21:29AM +0000, Patrick McHardy wrote:
> > >
> > > That doesn't look right, binops can also occur outside of relational
> > > expressions. I'd suggest to special case OP_EQ and not print it by
> > > default unless the LHS is an EXPR_BINOP.
> >
> > Something like this:
> >
> >
> > diff --git a/src/expression.c b/src/expression.c
> > index 71154cc..518f71c 100644
> > --- a/src/expression.c
> > +++ b/src/expression.c
> > @@ -356,7 +356,7 @@ const char *expr_op_symbols[] = {
> > [OP_XOR] = "^",
> > [OP_LSHIFT] = "<<",
> > [OP_RSHIFT] = ">>",
> > - [OP_EQ] = NULL,
> > + [OP_EQ] = "==",
> > [OP_NEQ] = "!=",
> > [OP_LT] = "<",
> > [OP_GT] = ">",
> > @@ -407,7 +407,9 @@ struct expr *unary_expr_alloc(const struct location *loc,
> > static void binop_expr_print(const struct expr *expr)
> > {
> > expr_print(expr->left);
> > - if (expr_op_symbols[expr->op] != NULL)
> > + if (expr_op_symbols[expr->op] &&
> > + (expr->op != OP_EQ ||
> > + expr->left->ops->type == EXPR_BINOP))
> > printf(" %s ", expr_op_symbols[expr->op]);
> > else
> > printf(" ");
>
> This looks a bit more complicated. To my understanding, the right-hand
> side of the relational tree contains the value. The left-hand side
> contains the binop tree, whose left-hand side is the meta mark and the
> right-hand side is the value to apply the operation. The print
> function doesn't have context to know what's on the right-hand side of
> the upper relational expression. Thinking how to fix this...
This OP_EQ case is the upper relational expression. Try it, it works fine :)
--
To unsubscribe from this list: send the line "unsubscribe netfilter-devel" in
the body of a message to majordomo@xxxxxxxxxxxxxxx
More majordomo info at http://vger.kernel.org/majordomo-info.html
