std::chrono::weekday::ok
来自cppreference.com
| |
(C++20 起) | |
检查存储于 *this 的星期之日值是否在有效范围，即 [0, 6] 中。
返回值
若存储于 *this 的星期之日值在范围 [0, 6] 中则为 true。否则为 false。
示例
运行此代码
#include <chrono>
#include <iomanip>
#include <iostream>
#include <locale>
#include <string>
struct weekday_ok : std::numpunct<char>
{
std::string do_truename() const override { return " (is valid weekday)"; }
std::string do_falsename() const override { return " (is not valid weekday)"; }
};
int main()
{
std::cout.imbue(std::locale(std::cout.getloc(), new weekday_ok));
std::cout << std::boolalpha;
for (const unsigned u : {0 /* Sun */, 1 /* Mon */, 6, 7 /* Sun */, 8, 9})
{
const std::chrono::weekday wd{u};
std::cout << "u: " << u << "; wd: " << wd.c_encoding() << wd.ok() << '\n';
}
}
输出：
u: 0; wd: 0 (is valid weekday)
u: 1; wd: 1 (is valid weekday)
u: 6; wd: 6 (is valid weekday)
u: 7; wd: 0 (is valid weekday)
u: 8; wd: 8 (is not valid weekday)
u: 9; wd: 9 (is not valid weekday)
参阅
| 取得存储的星期之日值 取得 ISO 8601 星期之日值 (公开成员函数) |
