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) |
