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Subject: Re: [boost] [ratio] metafunction ratio_subtract
From: Howard Hinnant (howard.hinnant_at_[hidden])
Date: 20091207 14:11:22
On Dec 7, 2009, at 6:42 AM, vicente.botet wrote:
>
>  Original Message 
> From: "Howard Hinnant" <howard.hinnant_at_[hidden]>
> To: <boost_at_[hidden]>
> Sent: Monday, December 07, 2009 12:18 AM
> Subject: Re: [boost] [ratio] metafunction ratio_subtract
>
>
>
>
> On Dec 6, 2009, at 3:43 PM, vicente.botet wrote:
>
>> Hi Beman, Hi Howard,
>>
>> With current implementation of the metafunction ratio_subtract I get the type from the test ratio_test.cpp
>>
>> User1::Distance d( User1::mile(110) );
>> User1::Time t( boost::chrono::hours(2) );
>>
>> typeof (d/t) is User1::quantity<boost::ratio<1,1>, boost::ratio<1,1> >
>>
>> So the following assignation do not works
>>
>> User1::Speed s = d / t;
>>
>> as User1::Speed is defined as
>>
>> typedef quantity<boost::ratio<1>, boost::ratio<0> > Time; // second
>> typedef quantity<boost::ratio<0>, boost::ratio<1> > Distance; // meter
>> typedef quantity<boost::ratio<1>, boost::ratio<1> > Speed; // meter/second
>>
>> The code is a little bit triky, and I have not taken the time to understand why the resulting type is
>> User1::quantity<boost::ratio<1,1>, boost::ratio<1,1> >
>> and not
>> <boost::ratio<1,1>, boost::ratio<1,1> > User1::quantity
>
> There must be a bug in the ratio formation of num and den. It should not be possible to form a ratio such that ratio::den is not positive. [ratio.ratio]/1 states that D shall not be 0. [ratio.ratio]/2 specifies that gcd operates on the absolute values of N and D (and so gcd must be positive). den is specified to be abs(D)/gcd, which must be positive.
>
>> I have replaced the type by
>>
>> typedef ratio<R1::num * R2::den  R2::num * R1::den,R1::den * R2::den> aux_type;
>> typedef ratio<aux_type::num ,aux_type::den > type; // get normalized type
>>
>> to follow the recommendation "The nested typedef type shall be a synonym for ratio<T1, T2> where T1 has the value R1::num *R2::den  R2::num * R1::den and T2 has the value R1::den * R2::den."
>>
>> and every thing is OK (See below the replaced code)
>>
>> I'm doing something wrong?
>
> No, you're doing something right by investigating this mystery. :)
>
> I've lost the link to the code you're working with. Could you post it?
>
> The original implementation of ratio_subtract (thank you for including it below) is designed to be equivalent to your substitution, but less susceptible to overflow, by performing exact division as soon as possible (get integers as low as possible, as soon as possible).
>
> I'm not sure where your example of computing speed from distance and time involves subtraction. But I'm interested to help you get to the bottom of it.
>
> Howard
> _______________________
> Hi,
>
> It is from the example related to Typesafe "physics" code interoperating with std::chrono::duration types and taking advantage of the std::ratio infrastructure and design philosophy. sandbox/chrono/libs/chrono/test/ratio_test.cpp
>
> The code I'm testing is derived from the one in the sandbox/chrono. I have modified something the implementation as the code didn't compiled on my environment (cygwin/gcc 3.4), most of them related to the fact that even on POSIX, the CLOCK_MONOTONIC is optionally defined. Please, could you check the unmodified code on the sandbox, if needed I can send you a patch this evening (french time).
Ah... Hadn't seen that code in awhile. :)
You've uncovered a design flaw in ratio. And your ratio<N, D>::type is the fix!
In terms of code:
1. type needs to be added to ratio:
typedef ratio<num, den> type;
2. ratio_multiply should typedef to ratio::type, instead of to ratio:
template <class _R1, class _R2>
struct ratio_multiply
{
private:
static const intmax_t __gcd_n1_d2 = __static_gcd<_R1::num, _R2::den>::value;
static const intmax_t __gcd_d1_n2 = __static_gcd<_R1::den, _R2::num>::value;
public:
typedef typename ratio
<
__ll_mul<_R1::num / __gcd_n1_d2, _R2::num / __gcd_d1_n2>::value,
__ll_mul<_R2::den / __gcd_n1_d2, _R1::den / __gcd_d1_n2>::value
>::type type;
};
3. Ditto for ratio_divide.
Points 2 and 3 will automatically fix ratio_add and ratio_subtract.
What is happening:
ratio<1, 1> was being given as the answer for ratio_subtract::type. This change gives instead ratio<1, 1>::type, which is ratio<1, 1>. It is not invalid to create a ratio with a negative denominator. But that ratio will be normalized to have a positive denominator, which is a different type. The bug was not recognizing this important distinction (my bad).
Howard
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