Re: [PATCH 2/2] expression: fix printing of binary operation

[Patrick McHardy <kaber@trash.net>]

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(" ");