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denc: Simplify is_any_of

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We can't simplify underlying type in the obvious way since template
evaluation is strict.

Signed-off-by: Adam C. Emerson <aemerson@redhat.com>
This commit is contained in:
Adam C. Emerson 2018-01-09 16:15:34 -05:00
parent 74948b5cd8
commit 617a0961ce
1 changed files with 7 additions and 12 deletions
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19
src/include/denc.h
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@ -54,6 +54,9 @@ struct denc_traits {
static constexpr bool need_contiguous = true;
};
template<typename T>
inline constexpr bool denc_supported = denc_traits<T>::supported;
// hack for debug only; FIXME
//#include <iostream>
@ -69,7 +72,7 @@ struct denc_traits {
//#define ENCODE_DUMP_PATH /tmp/something
#ifdef ENCODE_DUMP_PATH
# include <stdio.h>
# include <cstdio>
# include <sys/types.h>
# include <sys/stat.h>
# include <fcntl.h>
@ -224,16 +227,8 @@ struct denc_traits {
// ---------------------------------------------------------------------
// raw types
namespace _denc {
template<class T, class...>
struct is_any_of : std::false_type {};
template<class T, class Head, class... Tail>
struct is_any_of<T, Head, Tail...> : std::conditional<
std::is_same<T, Head>::value,
std::true_type,
is_any_of<T, Tail...>>::type
{};
template<typename T, typename... Us>
inline constexpr bool is_any_of_v = is_any_of<T, Us...>::value;
inline constexpr bool is_any_of = (... || std::is_same_v<T, Us>);
template<typename T, typename=void> struct underlying_type {
using type = T;
@ -251,8 +246,8 @@ template<typename T>
struct denc_traits<
T,
std::enable_if_t<
_denc::is_any_of_v<_denc::underlying_type_t<T>,
ceph_le64, ceph_le32, ceph_le16, uint8_t
_denc::is_any_of<_denc::underlying_type_t<T>,
ceph_le64, ceph_le32, ceph_le16, uint8_t
#ifndef _CHAR_IS_SIGNED
, int8_t
#endif