我做了一个模板函数,用于根据数字初始化chrono::time_point
。到目前为止,我已经成功了,但遇到了一个问题,我不完全理解。下面给出了我的代码的两个最小示例。
下面代码失败,错误如下进行编译:
/usr/include/c++/7/chrono:616:14: note: no known conversion for argument 1 from ‘const double’ to ‘const std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double> >&’
/usr/include/c++/7/chrono:616:14: note: candidate: constexpr std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double> >::time_point(std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double> >&&)
/usr/include/c++/7/chrono:616:14: note: no known conversion for argument 1 from ‘const double’ to ‘std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double> >&&’
#include <iostream>
#include <chrono>
namespace yv {
using clock_t = std::chrono::system_clock;
using duration_t = std::chrono::duration<double>;
using time_t = std::chrono::time_point<clock_t, duration_t>;
namespace fromnumber {
template<class T, class T_time> T_time time(T const& timestamp) {
return T_time(timestamp);
}
// No specialization
}; // end namespace fromnumber
}; // end namespace yv
int main()
{
using namespace yv;
using namespace std;
yv::time_t t0 = yv::fromnumber::time<double, yv::time_t>(0.0);
yv::time_t t1 = yv::fromnumber::time<double, yv::time_t>(1548675254.0);
return 0;
}
然而,当我添加模板特空定义它编译。</ P>
#include <iostream>
#include <chrono>
namespace yv {
using clock_t = std::chrono::system_clock;
using duration_t = std::chrono::duration<double>;
using time_t = std::chrono::time_point<clock_t, duration_t>;
namespace fromnumber {
template<class T, class T_time> T_time time(T const& timestamp) {
return T_time(timestamp);
}
template<> std::chrono::time_point<clock_t, duration_t> time(double const&) {
// EMPTY
}
}; // end namespace fromnumber
}; // end namespace yv
int main()
{
using namespace yv;
using namespace std;
yv::time_t t0 = yv::fromnumber::time<double, yv::time_t>(0.0);
yv::time_t t1 = yv::fromnumber::time<double, yv::time_t>(1548675254.0);
return 0;
}
在专业化具有定义,但它甚至不返回值。我在这里想念什么?
编辑: 感谢您的快速答复。以下是使用Howard Hinnant的date.h的更广泛示例。
#include <iostream>
#include "date/date.h"
//#include <chrono>
//using namespace date;
namespace yv {
using clock_t = std::chrono::system_clock;
using duration_t = std::chrono::duration<double>;
using time_t = std::chrono::time_point<clock_t, duration_t>;
namespace fromnumber {
template<class T, class T_time> T_time time(T const& timestamp) {
return T_time(timestamp);
}
// Case 1. Correct specialization, not getting any warnings.
template<> std::chrono::time_point<clock_t, duration_t> time(double const& t)
{
return std::chrono::time_point<clock_t, duration_t>(duration_t(t));
}
// Case 2. Incorrect specialization, compiles and prints the correct datetime but getting a warning
template<> std::chrono::time_point<clock_t, duration_t> time(double const& t)
{
}
// Case 3. Without the specialization it will not compile, error given above
}; // end namespace fromnumber
}; // end namespace yv
std::ostream& operator<< (std::ostream& outStream, const yv::time_t& t) {
using namespace date;
auto t2 = date::floor<std::chrono::milliseconds>(t);
outStream << date::format("%c", t2);
return outStream;
}
int main()
{
using namespace yv;
using namespace std;
yv::time_t t0 = yv::fromnumber::time<double, yv::time_t>(0.0);
yv::time_t t1 = yv::fromnumber::time<double, yv::time_t>(1548675254.0);
cout << t1 << endl;
// expecting: Mon Jan 28 11:34:14 2019
return 0;
}
第2种情况下的警告
../try_chrono/main.cpp: In function ‘T_time yv::fromnumber::time(const T&) [with T = double; T_time = std::chrono::time_point<std::chrono::_V2::system_clock, std::chrono::duration<double> >]’:
../try_chrono/main.cpp:21:1: warning: no return statement in function returning non-void [-Wreturn-type]
}
^
不从非void函数返回值显然是未定义的行为。不过,我不明白的事情是如何可能的是,我得到了正确的输出与空专业化? 我看到它的方式是,这两个壳2和3的情况下是不正确的,不应该给我在stdout正确的结果。
答案 0 :(得分:3)
未定义的行为。它会编译,但是您将收到一个巨大的警告,甚至可能是崩溃。还是什么都没有。
您知道的是您需要此定义,并且如定义所言,它应返回std::chrono::time_point<clock_t, duration_t>
。如果您不这样做,那您就违反了合同。编译器这样说:
warning: no return statement in function returning non-void [-Wreturn-type]
答案 1 :(得分:3)
模板专门化在应返回std::chrono::time_point<clock_t, duration_t>
时不返回任何内容,从而导致行为未定义。
标准在[stmt.return]/2中明确指出了这一点:
在没有return语句的情况下从值返回函数的末尾流出(main除外)是未定义的行为。
答案 2 :(得分:0)
通过与调试器我发现空定义返航编译后的参数的代码步进之后。编译器有效地改变了这一点:
template<> std::chrono::time_point<clock_t, duration_t> time(double t)
{
}
对此:
template<> std::chrono::time_point<clock_t, duration_t> time(double t)
{
return (std::chrono::time_point<clock_t, duration_t>) t;
}
这又导致了“正确的”二进制,因为类型std::chrono::time_point<clock_t, duration_t>
的实例在内存中看起来像这样:
name value address
t0 @0x0123456789ab
__d @0x0123456789ab
__r 1548675254.02 @0x0123456789ab
因此分配可以正确执行。
然而,没有返回参数此怪癖休息一个非空的专业化功能。
例如,以下函数不返回(std::chrono::time_point<clock_t, duration_t>) t
:
template<> std::chrono::time_point<clock_t, duration_t> time(double t)
{
cout << t << endl;
}
https://stackoverflow.com/a/1610454/2548426 根据该回答所得的二进制取决于平台,架构和编译器。
如先前的回答所述,这是未定义的行为。现在我很清楚是什么导致了明显的正确结果。
正确的专业化:
template<> std::chrono::time_point<clock_t, duration_t> time(double t)
{
return std::chrono::time_point<clock_t, duration_t>(duration_t(t));
}
或
template<> time_t time(double t)
{
return time_t(duration_t(t));
}