On Sun, May 18, 2025 at 11:57:15PM +0800, shejialuo wrote:
> When inserting an existing element, "add_entry" would convert "index"
> value to "-1-index" to indicate the caller that this element is in the
> list already.
>
> However, in "string_list_insert", we would simply convert this to the
> original positive index without any further action. Let's directly
> return the index as we don't care about whether the element is in the
> list by using "add_entry".
>
> In the future, if we want to let "add_entry" tell the caller, we may add
> "int *exact_match" parameter to "add_entry" instead of converting the
> index to negative to indicate.
>
> Signed-off-by: shejialuo <shejialuo@xxxxxxxxx>
> ---
> string-list.c | 6 +-----
> 1 file changed, 1 insertion(+), 5 deletions(-)
>
> diff --git a/string-list.c b/string-list.c
> index 8540c29bc9..171cef5dbb 100644
> --- a/string-list.c
> +++ b/string-list.c
> @@ -40,14 +40,13 @@ static int get_entry_index(const struct string_list *list, const char *string,
> return right;
> }
>
> -/* returns -1-index if already exists */
> static int add_entry(struct string_list *list, const char *string)
> {
> int exact_match = 0;
> int index = get_entry_index(list, string, &exact_match);
>
> if (exact_match)
> - return -1 - index;
> + return index;
>
> ALLOC_GROW(list->items, list->nr+1, list->alloc);
> if (index < list->nr)
Okay, let's assume that "index == 2" here and we have an exact match.
We'd thus return `-1 - 2 == -3`.
> @@ -65,9 +64,6 @@ struct string_list_item *string_list_insert(struct string_list *list, const char
> {
> int index = add_entry(list, string);
>
> - if (index < 0)
> - index = -1 - index;
> -
> return list->items + index;
> }
So we'd now realize that `index < 0` and thus calculate `-1 - -3 == 2`,
which is the original index indeed. So this is a nice simplification
that retains the original behaviour indeed.
I think we could simplify the code even further by inlining
`get_entry_index()` now that `string_list_insert()` is a trivial wrapper
around it. But I'll leave it up to you whether we want to do it or not.
Patrick
