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std::adjacent_difference

From cppreference.com
< cpp‎ | algorithm
 
 
 
Defined in header <numeric>
template< class InputIt, class OutputIt >

OutputIt adjacent_difference( InputIt first, InputIt last,

                              OutputIt d_first );
(1)
template< class InputIt, class OutputIt, class BinaryOperation >

OutputIt adjacent_difference( InputIt first, InputIt last,

                              OutputIt d_first, BinaryOperation op );
(2)

Computes the differences between the second and the first of each adjacent pair of elements of the range [first, last) and writes them to the range beginning at d_first + 1. Unmodified copy of first is written to d_first. The first version uses operator- to calculate the differences, the second version uses the given binary function op.

Equivalent operation:

*(d_first)   = *first;
*(d_first+1) = *(first+1) - *(first);
*(d_first+2) = *(first+2) - *(first+1);
*(d_first+3) = *(first+3) - *(first+2);
...

op must not have side effects.

(until C++11)

op must not invalidate any iterators, including the end iterators, or modify any elements of the ranges involved.

(since C++11)

Contents

[edit] Parameters

first, last - the range of elements
d_first - the beginning of the destination range
op - binary operation function object that will be applied.

The signature of the function should be equivalent to the following:

 Ret fun(const Type1 &a, const Type2 &b);

The signature does not need to have const &.
The types Type1 and Type2 must be such that an object of type iterator_traits<InputIt>::value_type can be implicitly converted to both of them. The type Ret must be such that an object of type OutputIt can be dereferenced and assigned a value of type Ret. ​

Type requirements
-
InputIt must meet the requirements of InputIterator.
-
OutputIt must meet the requirements of OutputIterator.

[edit] Return value

It to the element past the last element written.

[edit] Notes

If first == last, this function has no effect and will merely return d_first.

[edit] Complexity

Exactly (last - first) - 1 applications of the binary operation

[edit] Possible implementation

First version
template<class InputIt, class OutputIt>
OutputIt adjacent_difference(InputIt first, InputIt last, 
                             OutputIt d_first)
{
    if (first == last) return d_first;
 
    typedef typename std::iterator_traits<InputIt>::value_type value_t;
    value_t acc = *first;
    *d_first = acc;
    while (++first != last) {
        value_t val = *first;
        *++d_first = val - acc;
        acc = std::move(val);
    }
    return ++d_first;
}
Second version
template<class InputIt, class OutputIt, class BinaryOperation>
OutputIt adjacent_difference(InputIt first, InputIt last, 
                             OutputIt d_first, BinaryOperation op)
{
    if (first == last) return d_first;
 
    typedef typename std::iterator_traits<InputIt>::value_type value_t;
    value_t acc = *first;
    *d_first = acc;
    while (++first != last) {
        value_t val = *first;
        *++d_first = op(val, acc);
        acc = std::move(val);
    }
    return ++d_first;
}

[edit] Example

The following code converts a sequence of even numbers to repetitions of the number 2 and converts a sequence of ones to a sequence of Fibonacci numbers.

#include <numeric>
#include <vector>
#include <iostream>
#include <functional>
 
int main()
{
    std::vector<int> v{2, 4, 6, 8, 10, 12, 14, 16, 18, 20};
    std::adjacent_difference(v.begin(), v.end(), v.begin());
 
    for (auto n : v) {
        std::cout << n << ' ';
    }
    std::cout << '\n';
 
    v = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
    std::adjacent_difference(v.begin(), v.end() - 1, v.begin() + 1, std::plus<int>());
 
    for (auto n : v) {
        std::cout << n << ' ';
    }
    std::cout << '\n';
}

Output:

2 2 2 2 2 2 2 2 2 2
1 1 2 3 5 8 13 21 34 55

[edit] See also

computes the partial sum of a range of elements
(function template) [edit]
sums up a range of elements
(function template) [edit]
parallelized version of std::ajacent_difference
(function template) [edit]