c++ - 从time_point +持续时间到持续时间的转换在Clang上失败

此代码在gcc-4.8上编译，而在clang-3.3上失败？
以及如何使此代码在clang上可编译？ = \

#include <chrono>
#include <iostream>
#include <thread>

void sleep_until_wow(const std::chrono::system_clock::time_point& time)
{
    std::cout << "zzz...\n";
    std::this_thread::sleep_until(time);
    std::cout << "wake up!\n";
}

template <typename Rep, typename Period>
void sleep_for(const std::chrono::duration<Rep, Period>& duration)
{
    sleep_until_wow(std::chrono::system_clock::now() + duration);
}

template <typename Clock, typename Duration>
void sleep_until(const std::chrono::time_point<Clock, Duration>& time)
{
    sleep_until_wow(time);
}


int main() {
    sleep_for(std::chrono::nanoseconds(500000));
    return 0;
}

错误信息：

% clang++ -std=c++11 -stdlib=libc++ time_point.cpp
time_point.cpp:17:5: error: no matching function for call to 'sleep_until_wow'
    sleep_until_wow(std::chrono::system_clock::now() + ns);
    ^~~~~~~~~~~~~~~
time_point.cpp:28:5: note: in instantiation of function template specialization 'sleep_for<long long, std::__1::ratio<1, 1000000000> >' requested here
    sleep_for(std::chrono::nanoseconds(500000));
    ^
time_point.cpp:5:6: note: candidate function not viable: no known conversion from 'time_point<[...], typename common_type<class duration<long long, class ratio<1, 1000000> >, duration<long long, class ratio<1, 1000000000> >
      >::type>' to 'const time_point<[...], (default) typename _Clock::duration>' for 1st argument
void sleep_until_wow(const std::chrono::system_clock::time_point& time)
     ^
1 error generated.

最佳答案

好吧，我明白了。

在gcc stl中，system_clock的持续时间为纳秒，但在clang中为微秒。

关于c++ - 从time_point +持续时间到持续时间的转换在Clang上失败，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/22205290/