From mboxrd@z Thu Jan  1 00:00:00 1970
From: Pablo Neira Ayuso <pablo@netfilter.org>
Subject: Re: [PATCH 2/2] expression: fix printing of binary operation
Date: Wed, 15 Jan 2014 12:41:34 +0100
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Cc: netfilter-devel@vger.kernel.org
To: Patrick McHardy <kaber@trash.net>
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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:
> > On Wed, Jan 15, 2014 at 12:09:27PM +0100, Pablo Neira Ayuso wrote:
> > > This patch adds a special print function for the relational case in
> > > which == is assumed, so it's not printed. It also fixes the output of
> > > binary operations from:
> > > 
> > > & 0x00000003 0x00000001
> > > 
> > > to:
> > > 
> > > and 0x00000003 == 0x00000001
> > > 
> > > diff --git a/src/expression.c b/src/expression.c
> > > index 6da5c10..452b0d7 100644
> > > --- a/src/expression.c
> > > +++ b/src/expression.c
> > > @@ -411,7 +411,9 @@ static void binop_expr_print(const struct expr *expr)
> > >  		printf(" %s ", expr_op_symbols[expr->op]);
> > >  	else
> > >  		printf(" ");
> > > +
> > >  	expr_print(expr->right);
> > > +	printf(" ==");
> > 
> > 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...