std::ratio_add
Defined in header <ratio>
|
||
template< class R1, class R2 > using ratio_add = /* see below */; |
(since C++11) | |
The alias template std::ratio_add
denotes the result of adding 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.
[edit] 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_add<R1, R2> be already reduced to lowest terms; for example, std::ratio_add<std::ratio<1, 3>, std::ratio<1, 6>> is the same type as std::ratio<1, 2>.
[edit] Example
#include <iostream> #include <ratio> int main() { using two_third = std::ratio<2, 3>; using one_sixth = std::ratio<1, 6>; using sum = std::ratio_add<two_third, one_sixth>; std::cout << "2/3 + 1/6 = " << sum::num << '/' << sum::den << '\n'; }
Output:
2/3 + 1/6 = 5/6
[edit] See also
(C++11) |
subtracts two ratio objects at compile-time(alias template) |