On Jan 5, 2012, at 10:30 AM, ivan.lelann@free.fr wrote:
Hi,
As of 1.48 Boost.Chrono contains code below for boost::chrono::process_real_cpu_clock::now() (boost\chrono\detail\inlined\win\process_cpu_clocks.hpp)
clock_t c = ::clock(); /* ... */ return time_point( duration(c*(1000000000l/CLOCKS_PER_SEC)) );
duration::rep is int64/nanoseconds, clock_t is long. This is VS2008, Win XP 32bits.
I think "c" should be cast to duration::rep before being multiplied. I randomly get negative time_point and this seems to be fixed by replacing "c" with "((duration::rep) (c))" in duration constructor :
duration( ((duration::rep) (c)) *(1000000000l/CLOCKS_PER_SEC))
Can someone confirm my view and that the fix is correct ?
Your fix looks correct to me. I'm not positive what value CLOCKS_PER_SEC has on your platform. But here is a reformulation of your solution that might be a little safer (depending on what CLOCKS_PER_SEC actually is): typedef ratio_divide<giga, ratio<CLOCKS_PER_SEC>>::type R; return time_point( duration(static_cast<rep>(c)*R::num/R::den) ); Using giga avoids mistakes in counting zeros. Using ratio_divide reduces the fraction 1000000000l/CLOCKS_PER_SEC to lowest terms, no matter the value of CLOCKS_PER_SEC. You can use rep as a shortcut for duration::rep, assuming this is inside of the clock's now() function. static_cast<rep> is a little safer than a C cast. I would expect the results to be the same, but it is good to code defensively. If you know a-priori that R::den is 1, you might: typedef ratio_divide<giga, ratio<CLOCKS_PER_SEC>>::type R; static_assert(R::den == 1, "oops!"); return time_point( duration(static_cast<rep>(c)*R::num) ); Howard