On 08/08/2019 13:31, Krzysztof Kozlowski wrote:
On Thu, 8 Aug 2019 at 14:07, Sylwester Nawrocki <s.nawrocki@xxxxxxxxxxx> wrote:
+static unsigned int exynos5422_asv_parse_table(struct exynos_asv *asv,
+ unsigned int pkg_id)
+{
+ return (pkg_id >> EXYNOS5422_TABLE_OFFSET) & EXYNOS5422_TABLE_MASK;
+}
+
+static bool exynos5422_asv_parse_bin2(struct exynos_asv *asv,
+ unsigned int pkg_id)
+{
+ return (pkg_id >> EXYNOS5422_BIN2_OFFSET) & EXYNOS5422_BIN2_MASK;
return !!() for converting to boolean.
I'm not convinced it is needed, the return type of the function is bool
and value of the expression will be implicitly converted to that type.
Is there any compiler warning related to that?
Yeah, but bool is int so there will be no implicit conversion... I
guess it is a convention. In theory !! is the proper conversion to
bool but if bool==int then it's essentially conversion to 1. I am not
sure what's the benefit, maybe for some wrong code which would do
comparisons on result like if (exynos5422_asv_parse_bin2() == TRUE)...
Not so - since we use "-std=gnu89", we have C99-like _Bool, which our
bool is a typedef of. Conversions, either implicit or explicit, are
well-defined:
"6.3.1.2 Boolean type
When any scalar value is converted to _Bool, the result is 0 if the
value compares equal
to 0; otherwise, the result is 1."
This is even called out in Documentation/process/coding-style.rst:
"When using bool types the !! construction is not needed, which
eliminates a class of bugs."
Robin.
