我有一个静态的结构数组:
struct CommandStruct
{
char* data;
unsigned ans_size;
};
static const CommandStruct commands[] =
{
{ "Some literal", 28 },
{ "Some other literal", 29 },
{ "Yet another literal", 8 },
};
我希望字符串是16字节对齐的。有可能直接实现吗?
我可能会单独定义每个文字,比如__declspec(align(16)) static const char some_command_id[] = "my literal"
,但那是一团糟。我需要在一个代码块中进行所有初始化。
答案 0 :(得分:2)
使用C ++ 11,以下内容可能有所帮助:https://ideone.com/IDEdY0
#include <cstdint>
// Sequence of char
template <char...Cs> struct char_sequence
{
template <char C> using push_back = char_sequence<Cs..., C>;
};
// Remove all chars from char_sequence from '\0'
template <typename, char...> struct strip_sequence;
template <char...Cs>
struct strip_sequence<char_sequence<>, Cs...>
{
using type = char_sequence<Cs...>;
};
template <char...Cs, char...Cs2>
struct strip_sequence<char_sequence<'\0', Cs...>, Cs2...>
{
using type = char_sequence<Cs2...>;
};
template <char...Cs, char C, char...Cs2>
struct strip_sequence<char_sequence<C, Cs...>, Cs2...>
{
using type = typename strip_sequence<char_sequence<Cs...>, Cs2..., C>::type;
};
// struct to create a aligned char array
template <std::size_t Alignment, typename chars> struct aligned_string;
template <std::size_t Alignment, char...Cs>
struct aligned_string<Alignment, char_sequence<Cs...>>
{
alignas(Alignment) static constexpr char str[sizeof...(Cs)] = {Cs...};
};
template <std::size_t Alignment, char...Cs>
alignas(Alignment) constexpr
char aligned_string<Alignment, char_sequence<Cs...>>::str[sizeof...(Cs)];
// helper to get the i_th character (`\0` for out of bound)
template <std::size_t I, std::size_t N>
constexpr char at(const char (&a)[N]) { return I < N ? a[I] : '\0'; }
// helper to check if the c-string will not be truncated
template <std::size_t max_size, std::size_t N>
constexpr bool check_size(const char (&)[N])
{
static_assert(N <= max_size, "string too long");
return N <= max_size;
}
// Helper macros to build char_sequence from c-string
#define PUSH_BACK_8(S, I) \
::push_back<at<(I) + 0>(S)>::push_back<at<(I) + 1>(S)> \
::push_back<at<(I) + 2>(S)>::push_back<at<(I) + 3>(S)> \
::push_back<at<(I) + 4>(S)>::push_back<at<(I) + 5>(S)> \
::push_back<at<(I) + 6>(S)>::push_back<at<(I) + 7>(S)>
#define PUSH_BACK_32(S, I) \
PUSH_BACK_8(S, (I) + 0) PUSH_BACK_8(S, (I) + 8) \
PUSH_BACK_8(S, (I) + 16) PUSH_BACK_8(S, (I) + 24)
#define PUSH_BACK_128(S, I) \
PUSH_BACK_32(S, (I) + 0) PUSH_BACK_32(S, (I) + 32) \
PUSH_BACK_32(S, (I) + 64) PUSH_BACK_32(S, (I) + 96)
// Macro to create char_sequence from c-string (limited to 128 chars)
#define MAKE_CHAR_SEQUENCE(S) \
strip_sequence<char_sequence<> \
PUSH_BACK_128(S, 0) \
>::type::template push_back<check_size<128>(S) ? '\0' : '\0'>
// Macro to return an aligned c-string
#define MAKE_ALIGNED_STRING(ALIGNMENT, S) \
aligned_string<ALIGNMENT, MAKE_CHAR_SEQUENCE(S)>::str
所以你有:
static const CommandStruct commands[] =
{
{ MAKE_ALIGNED_STRING(16, "Some literal"), 28 },
{ MAKE_ALIGNED_STRING(16, "Some other literal"), 29 },
{ MAKE_ALIGNED_STRING(16, "Yet another literal"), 8 },
};
答案 1 :(得分:0)
浏览过一段时间后,我已经设法做了一些东西,它只是自动化单独的文字构造,(我也会对所有数组的元素进行枚举):
#define CMD_TUPLE ( \
(cmdCommandOne, "The first command", 1900),\
(cmdCommandTwo, "The second one", 1),\
(cmdAnother, "Another command", 11))
#define CMD_SEQ BOOST_PP_TUPLE_TO_SEQ(CMD_TUPLE)
#define CMD_MAKE_ENUM(r, data, elem) BOOST_PP_TUPLE_ELEM(0, elem),
enum Commands { BOOST_PP_SEQ_FOR_EACH(CMD_MAKE_ENUM, , CMD_SEQ) cmdLast };
#define CMD_MAKE_STRING(r, data, elem) \
__declspec(align(16)) static const char \
BOOST_PP_CAT(cmd_, BOOST_PP_TUPLE_ELEM(0, elem))[] = BOOST_PP_TUPLE_ELEM(1, elem);
BOOST_PP_SEQ_FOR_EACH(CMD_MAKE_STRING, , CMD_SEQ)
#define CMD_MAKE_ARRAY(r, data, elem) \
{ BOOST_PP_CAT(cmd_, BOOST_PP_TUPLE_ELEM(0, elem)), \
BOOST_PP_SEQ_ENUM(BOOST_PP_SEQ_REST_N(2, BOOST_PP_TUPLE_TO_SEQ(elem))) },
static const CommandStruct commands[] = {
BOOST_PP_SEQ_FOR_EACH(CMD_MAKE_ARRAY, , CMD_SEQ)
};