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Difference between revisions of "cpp/numeric/ratio/ratio subtract"

From cppreference.com
< cpp‎ | numeric‎ | ratio
m (Shorten template names. Use {{lc}} where appropriate.)
(Fix wording)
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{{dcl | 1=
 
{{dcl | 1=
 
template< class R1, class R2 >
 
template< class R1, class R2 >
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using ratio_subtract = /* unspecified */;
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using ratio_subtract = /* see below */;
 
}}
 
}}
 
{{dcl end}}
 
{{dcl end}}
  
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The template alias {{tt|std::ratio_subtract}} denotes the result of subtracting two exact rational fractions represented by the {{lc|std::ratio}} instances {{tt|R1}} and {{tt|R2}}. The result a {{lc|std::ratio}} instance {{tt|std::ratio<Num, Denom>}} where {{c|1=Num == R1::num * R2::den - R2::num * R1::den}} and {{c|1=Denom == R1::den * R2::den}}.
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The alias template {{tt|std::ratio_subtract}} denotes the result of subtracting two exact rational fractions represented by the {{lc|std::ratio}} specializations {{tt|R1}} and {{tt|R2}}.
  
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===Member types===
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The result is a {{lc|std::ratio}} specialization {{c|std::ratio<U, V>}}, such that given {{c|1=Num == R1::num * R2::den - R2::num * R1::den}} and {{c|1=Denom == R1::den * R2::den}} (computed without arithmetic overflow), {{tt|U}} is {{c|std::ratio<Num, Denom>::num}} and {{tt|V}} is {{c|std::ratio<Num, Denom>::den}}.
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{{dsc begin}}
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{{dsc hitem | Member type | Definition}}
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{{dsc | {{tt|type}} | {{c|std::ratio<num, den>}}}}
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{{dsc end}}
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===Member constants===
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===Notes===
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{{dsc begin}}
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If {{tt|U}} or {{tt|V}} is not representable in {{tt|std::intmax_t}}, the program is ill-formed. If {{tt|Num}} or {{tt|Denom}} is not representable in {{tt|std::intmax_t}}, the program is ill-formed unless the implementation yields correct values for {{tt|U}} and {{tt|V}}.
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{{dsc mem sconst | nolink=true | num | {{c|constexpr}} value of type {{lc|std::intmax_t}} equal to {{tt|sign(Num) * sign(Denom) * abs(Num) / gcd(Num, Denom)}} }}
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{{dsc mem sconst | nolink=true | den | {{c|constexpr}} value of type {{lc|std::intmax_t}} equal to {{tt|abs(Denom) / gcd(Num, Denom)}} }}
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The above definition requires that the result of {{c|std::ratio_subtract<R1, R2>}} be already reduced to lowest terms; for example, {{c|std::ratio_subtract<std::ratio<1, 2>, std::ratio<1, 6>>}} is the same type as {{c|std::ratio<1, 3>}}.
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{{dsc end}}
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===Example===
 
===Example===

Revision as of 15:22, 23 May 2015

 
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Defined in header <ratio>
template< class R1, class R2 >
using ratio_subtract = /* see below */;

The alias template std::ratio_subtract denotes the result of subtracting two exact rational fractions represented by the std::ratio specializations R1 and R2.

The result is a std::ratio specialization std::ratio<U, V>, such that given Num == R1::num * R2::den - R2::num * R1::den and Denom == R1::den * R2::den (computed without arithmetic overflow), U is std::ratio<Num, Denom>::num and V is std::ratio<Num, Denom>::den.

Notes

If U or V is not representable in std::intmax_t, the program is ill-formed. If Num or Denom is not representable in std::intmax_t, the program is ill-formed unless the implementation yields correct values for U and V.

The above definition requires that the result of std::ratio_subtract<R1, R2> be already reduced to lowest terms; for example, std::ratio_subtract<std::ratio<1, 2>, std::ratio<1, 6>> is the same type as std::ratio<1, 3>.

Example

Run this code
#include <iostream>
#include <ratio>
 
int main()
{
    typedef std::ratio<2, 3> two_third;
    typedef std::ratio<1, 6> one_sixth;
 
    typedef std::ratio_subtract<two_third, one_sixth> diff;
    std::cout << "2/3 - 1/6 = " << diff::num << '/' << diff::den << '\n';
}

Output: 

2/3 - 1/6 = 1/2
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