Re: [PATCH v3 3/9] rbtree: place easiest case first in rb_erase()

[Andrew Morton]

On Mon, 20 Aug 2012 15:05:25 -0700
Michel Lespinasse <walken@xxxxxxxxxx> wrote:

> In rb_erase, move the easy case (node to erase has no more than
> 1 child) first. I feel the code reads easier that way.

Well.  For efficiency we should put the commonest case first.  Is that
the case here?

> --- a/lib/rbtree.c
> +++ b/lib/rbtree.c
> @@ -368,17 +368,28 @@ static void __rb_erase_color(struct rb_node *node, struct rb_node *parent,
>  
>  void rb_erase(struct rb_node *node, struct rb_root *root)
>  {
> -	struct rb_node *child, *parent;
> +	struct rb_node *child = node->rb_right, *tmp = node->rb_left;

Coding style nit: multiple-definitions-per-line makes it harder to
locate a particular definition, and from long experience I can assure
you that it makes management of subsequent overlapping patches quite a
lot harder.  Also, one-definition-per-line gives room for a nice little
comment, and we all like nice little comments.

Also, "tmp" is a rotten name.  Your choice of an identifier is your
opportunity to communicate something to the reader.  When you choose
"tmp", you threw away that opportunity.  Should it be called "left"?


--- a/lib/rbtree.c~rbtree-place-easiest-case-first-in-rb_erase-fix
+++ a/lib/rbtree.c
@@ -368,12 +368,13 @@ static void __rb_erase_color(struct rb_n
 
 void rb_erase(struct rb_node *node, struct rb_root *root)
 {
-	struct rb_node *child = node->rb_right, *tmp = node->rb_left;
+	struct rb_node *child = node->rb_right
+	struct rb_node *tmp = node->rb_left;
 	struct rb_node *parent;
 	int color;
 
 	if (!tmp) {
-	case1:
+case1:
 		/* Case 1: node to erase has no more than 1 child (easy!) */
 
 		parent = rb_parent(node);
_

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