std::reduce
|   Defined in header  <numeric>
  | 
||
| (1) | ||
|   template<class InputIt> typename std::iterator_traits<InputIt>::value_type reduce(  | 
 (since C++17)  (until C++20)  | 
|
|   template<class InputIt> constexpr typename std::iterator_traits<InputIt>::value_type reduce(  | 
(since C++20) | |
|   template<class ExecutionPolicy, class ForwardIt> typename std::iterator_traits<ForwardIt>::value_type reduce(  | 
(2) | (since C++17) | 
| (3) | ||
|   template<class InputIt, class T> T reduce(InputIt first, InputIt last, T init);  | 
 (since C++17)  (until C++20)  | 
|
|   template<class InputIt, class T> constexpr T reduce(InputIt first, InputIt last, T init);  | 
(since C++20) | |
|   template<class ExecutionPolicy, class ForwardIt, class T> T reduce(ExecutionPolicy&& policy,  | 
(4) | (since C++17) | 
| (5) | ||
|   template<class InputIt, class T, class BinaryOp> T reduce(InputIt first, InputIt last, T init, BinaryOp binary_op);  | 
 (since C++17)  (until C++20)  | 
|
|   template<class InputIt, class T, class BinaryOp> constexpr T reduce(InputIt first, InputIt last, T init, BinaryOp binary_op);  | 
(since C++20) | |
|   template<class ExecutionPolicy, class ForwardIt, class T, class BinaryOp> T reduce(ExecutionPolicy&& policy,  | 
(6) | (since C++17) | 
init over binary_op. policy. These overloads do not participate in overload resolution unless  std::is_execution_policy_v<std::decay_t<ExecutionPolicy>>  (until C++20)  std::is_execution_policy_v<std::remove_cvref_t<ExecutionPolicy>>  (since C++20) is true.The behavior is non-deterministic if binary_op is not associative or not commutative.
The behavior is undefined if binary_op modifies any element or invalidates any iterator in [first; last], including the end iterator.
Contents | 
[edit] Parameters
| first, last | - | the range of elements to apply the algorithm to | 
| init | - | the initial value of the generalized sum | 
| policy | - | the execution policy to use. See execution policy for details. | 
| binary_op | - |   binary FunctionObject that will be applied in unspecified order to the result of dereferencing the input iterators, the results of other binary_op and init.
 | 
| Type requirements | ||
 -InputIt must meet the requirements of LegacyInputIterator.
 | ||
 -ForwardIt must meet the requirements of LegacyForwardIterator.
 | ||
 -T must meet the requirements of MoveConstructible. and binary_op(init, *first), binary_op(*first, init), binary_op(init, init), and binary_op(*first, *first) must be convertible to T.
 | ||
[edit] Return value
Generalized sum of init and *first, *(first+1), ... *(last-1) over binary_op,
where generalized sum GSUM(op, a
1, ..., a
N) is defined as follows: 
-  if N=1, a
1 -  if N > 1, op(GSUM(op, b
1, ..., b
K), GSUM(op, b
M, ..., b
N)) where 
-  b
1, ..., b
N may be any permutation of a1, ..., aN and - 1 < K+1 = M ≤ N
 
-  b
 
in other words, reduce behaves like std::accumulate except the elements of the range may be grouped and rearranged in arbitrary order
[edit] Complexity
O(last - first) applications of binary_op.
[edit] Exceptions
The overloads with a template parameter named ExecutionPolicy report errors as follows:
-  If execution of a function invoked as part of the algorithm throws an exception and 
ExecutionPolicyis one of the standard policies, std::terminate is called. For any otherExecutionPolicy, the behavior is implementation-defined. - If the algorithm fails to allocate memory, std::bad_alloc is thrown.
 
[edit] Notes
If the range is empty, init is returned, unmodified
[edit] Example
side-by-side comparison between reduce and std::accumulate:
#include <chrono> #include <execution> #include <iomanip> #include <iostream> #include <numeric> #include <utility> #include <vector> int main() { auto eval = [](auto fun) { const auto t1 = std::chrono::high_resolution_clock::now(); const auto [name, result] = fun(); const auto t2 = std::chrono::high_resolution_clock::now(); const std::chrono::duration<double, std::milli> ms = t2 - t1; std::cout << std::fixed << std::setprecision(1) << name << " result " << result << " took " << ms.count() << " ms\n"; }; { const std::vector<double> v(100'000'007, 0.1); eval([&v]{ return std::pair{"std::accumulate (double)", std::accumulate(v.cbegin(), v.cend(), 0.0)}; } ); eval([&v]{ return std::pair{"std::reduce (seq, double)", std::reduce(std::execution::seq, v.cbegin(), v.cend())}; } ); eval([&v]{ return std::pair{"std::reduce (par, double)", std::reduce(std::execution::par, v.cbegin(), v.cend())}; } ); }{ const std::vector<long> v(100'000'007, 1); eval([&v]{ return std::pair{"std::accumulate (long)", std::accumulate(v.cbegin(), v.cend(), 0)}; } ); eval([&v]{ return std::pair{"std::reduce (seq, long)", std::reduce(std::execution::seq, v.cbegin(), v.cend())}; } ); eval([&v]{ return std::pair{"std::reduce (par, long)", std::reduce(std::execution::par, v.cbegin(), v.cend())}; } ); } }
Possible output:
std::accumulate (double) result 10000000.7 took 163.6 ms std::reduce (seq, double) result 10000000.7 took 162.9 ms std::reduce (par, double) result 10000000.7 took 97.5 ms std::accumulate (long) result 100000007 took 62.3 ms std::reduce (seq, long) result 100000007 took 64.3 ms std::reduce (par, long) result 100000007 took 49.0 ms
[edit] See also
|    sums up a range of elements   (function template)  | |
|    applies a function to a range of elements, storing results in a destination range   (function template)  | |
|    (C++17)  | 
   applies an invocable, then reduces out of order   (function template)  |