std::reduce (3) - Linux Manuals

std::reduce: std::reduce

NAME

std::reduce - std::reduce

Synopsis

Defined in header <numeric>
template<class InputIt>
typename std::iterator_traits<InputIt>::value_type reduce( (1) (since C++17)
InputIt first, InputIt last);
template<class ExecutionPolicy, class ForwardIt>
typename std::iterator_traits<ForwardIt>::value_type reduce( (2) (since C++17)
ExecutionPolicy&& policy,
ForwardIt first, ForwardIt last);
template<class InputIt, class T> (3) (since C++17)
T reduce(InputIt first, InputIt last, T init);
template<class ExecutionPolicy, class ForwardIt, class T>
T reduce(ExecutionPolicy&& policy, (4) (since C++17)
ForwardIt first, ForwardIt last, T init);
template<class InputIt, class T, class BinaryOp> (5) (since C++17)
T reduce(InputIt first, InputIt last, T init, BinaryOp binary_op);
template<class ExecutionPolicy, class ForwardIt, class T, class BinaryOp>
T reduce(ExecutionPolicy&& policy, (6) (since C++17)
ForwardIt first, ForwardIt last, T init, BinaryOp binary_op);

1) same as reduce(first, last, typename std::iterator_traits<InputIt>::value_type{})
3) same as reduce(first, last, init, std::plus<>())
5) Reduces the range [first; last), possibly permuted and aggregated in unspecified manner, along with the initial value init over binary_op.
2,4,6) Same as (1,3,5), but executed according to policy. This overload only participates in overload resolution if std::is_execution_policy_v<std::decay_t<ExecutionPolicy>> 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.

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.

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

in other words, reduce behaves like std::accumulate except the elements of the range may be grouped and rearranged in arbitrary order

Complexity

O(last - first) applications of binary_op.

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 ExecutionPolicy is one of the standard_policies, std::terminate is called. For any other ExecutionPolicy, the behavior is implementation-defined.
* If the algorithm fails to allocate memory, std::bad_alloc is thrown.

Notes

If the range is empty, init is returned, unmodified

Example

side-by-side comparison between reduce and std::accumulate:
// Run this code

  #include <iostream>
  #include <chrono>
  #include <vector>
  #include <numeric>
  #include <execution>

  int main()
  {
      std::vector<double> v(10'000'007, 0.5);

      {
          auto t1 = std::chrono::high_resolution_clock::now();
          double result = std::accumulate(v.begin(), v.end(), 0.0);
          auto t2 = std::chrono::high_resolution_clock::now();
          std::chrono::duration<double, std::milli> ms = t2 - t1;
          std::cout << std::fixed << "std::accumulate result " << result
                    << " took " << ms.count() << " ms\n";
      }

      {
          auto t1 = std::chrono::high_resolution_clock::now();
          double result = std::reduce(std::execution::par, v.begin(), v.end());
          auto t2 = std::chrono::high_resolution_clock::now();
          std::chrono::duration<double, std::milli> ms = t2 - t1;
          std::cout << "std::reduce result "
                    << result << " took " << ms.count() << " ms\n";
      }
  }

Possible output:

  std::accumulate result 5000003.50000 took 12.7365 ms
  std::reduce result 5000003.50000 took 5.06423 ms