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c++ - std::chrono::years 存储真的至少是 17 位吗?

转载 作者:行者123 更新时间:2023-12-01 09:19:30 24 4
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来自 cppreference

std::chrono::years (since C++20)
duration</*signed integer type of at least 17 bits*/, std::ratio<31556952>>



使用 libc++ ,似乎是 std::chrono::years的下划线存储是 short签名 16 位 .
std::chrono::years( 30797 )        // yields  32767/01/01
std::chrono::years( 30797 ) + 365d // yields -32768/01/01 apparently UB

cppreference 上是否有错别字?还是别的什么?

例子:
#include <fmt/format.h>
#include <chrono>

template <>
struct fmt::formatter<std::chrono::year_month_day> {
char presentation = 'F';

constexpr auto parse(format_parse_context& ctx) {
auto it = ctx.begin(), end = ctx.end();
if (it != end && *it == 'F') presentation = *it++;

# ifdef __exception
if (it != end && *it != '}') {
throw format_error("invalid format");
}
# endif

return it;
}

template <typename FormatContext>
auto format(const std::chrono::year_month_day& ymd, FormatContext& ctx) {
int year(ymd.year() );
unsigned month(ymd.month() );
unsigned day(ymd.day() );
return format_to(
ctx.out(),
"{:#6}/{:#02}/{:#02}",
year, month, day);
}
};

using days = std::chrono::duration<int32_t, std::ratio<86400> >;
using sys_day = std::chrono::time_point<std::chrono::system_clock, std::chrono::duration<int32_t, std::ratio<86400> >>;

template<typename D>
using sys_time = std::chrono::time_point<std::chrono::system_clock, D>;
using sys_day2 = sys_time<days>;

int main()
{
auto a = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::hours( (1<<23) - 1 )
)
)
);

auto b = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::minutes( (1l<<29) - 1 )
)
)
);

auto c = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::seconds( (1l<<35) - 1 )
)
)
);

auto e = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::days( (1<<25) - 1 )
)
)
);

auto f = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::weeks( (1<<22) - 1 )
)
)
);

auto g = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::months( (1<<20) - 1 )
)
)
);

auto h = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::years( 30797 ) // 0x7FFF - 1970
)
)
);

auto i = std::chrono::year_month_day(
sys_day(
std::chrono::floor<days>(
std::chrono::years( 30797 ) // 0x7FFF - 1970
) + std::chrono::days(365)
)
);

fmt::print("Calendar limit by duration's underlining storage:\n"
"23 bit hour : {:F}\n"
"29 bit minute : {:F}\n"
"35 bit second : {:F}\n"
"25 bit days : {:F}\n"
"22 bit week : {:F}\n"
"20 bit month : {:F}\n"
"16? bit year : {:F}\n"
"16? bit year+365d : {:F}\n"
, a, b, c, e, f, g, h, i);
}

[ Godbolt link ]

最佳答案

cppreference 文章是 correct .如果 libc++ 使用较小的类型,那么这似乎是 libc++ 中的错误。

关于c++ - std::chrono::years 存储真的至少是 17 位吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/60668404/

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