Std::reduce (3) Linux Manual Page

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

std::reduce (3) Linux Manual Page