accumulate()

Library:  Numerics

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A generalized numeric operation that accumulates all elements within a range into a single value

#include <numeric>

namespace std {
  template <class InputIterator, class T>
  T accumulate(InputIterator start,
               InputIterator finish,
               T init);

  template <class InputIterator,
            class T,
            class BinaryOperation>
  T accumulate(InputIterator start,
               InputIterator finish,
               T init,
               BinaryOperation binary_op);
}

accumulate() applies a binary operation to init and each value in the range [start,finish). The result of each operation is returned in init. This process aggregates the result of performing the operation on every element of the sequence into a single value.

The accumulator acc is initialized with the value init and modified with acc = acc + *i or acc = binary_op(acc, *i) for each interator i, in order, in the range [start, finish).

accumulate() performs exactly finish-start applications of the binary operation, operator+ by default.

//
// accum.cpp
//

#include <numeric>    // for accumulate
#include <vector>     // for vector
#include <functional> // for multiplies
#include <iostream>   // for cout


int main ()
{
    // Typedef for convenience.
    typedef std::vector<int, std::allocator<int> > vector;

    // Initialize a vector using an array of integers.
    const vector::value_type arr[] = 
      { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
    vector v1 (arr, arr + sizeof arr / sizeof *arr);

    // Accumulate sums and products.
    vector::value_type sum =
        std::accumulate (v1.begin (), v1.end (), 0);

    vector::value_type prod =
        std::accumulate (v1.begin (), v1.end (), 1,
                         std::multiplies<vector::value_type>());

    // Output the results.
    std::cout << "For the series: ";
    for (vector::iterator i = v1.begin (); i != v1.end (); ++i)
        std::cout << *i << " ";

    std::cout << "where N = " << v1.size ()
              << "\nThe sum = (N * N + N) / 2 = " << sum
              << "\nThe product = N! = " << prod << std::endl;

    return 0;
}


Program Output:

For the series: 1 2 3 4 5 6 7 8 9 10 where N = 10
The sum = (N * N + N) / 2 = 55
The product = N! = 3628800

ISO/IEC 14882:1998 -- International Standard for Information Systems -- Programming Language C++, Section 26.4.1