std::accumulate() algorithm
- od C++20
- do C++20
// (1)
template< class InputIt, class T >
constexpr T accumulate( InputIt first, InputIt last, T init );
// (2)
template< class InputIt, class T, class BinaryOperation >
constexpr T accumulate( InputIt first, InputIt last, T init, BinaryOperation op );
// (1)
template< class InputIt, class T >
T accumulate( InputIt first, InputIt last, T init );
// (2)
template< class InputIt, class T, class BinaryOperation >
T accumulate( InputIt first, InputIt last, T init, BinaryOperation op );
Computes the sum of the given value init
and the elements in the ange [first
; last
).
Initializes the accumulator acc
(of type T
) with the initial value init
and then modifies it with:
- (1)
acc = acc + *i
(do C++20)acc = std::move(acc) + *i
(od C++20) - (2)
acc = op(acc, *i)
(do C++20)acc = op(std::move(acc), *i)
(od C++20)
for every iterator i
in the range [first
; last
) in order.
If op
invalidates any iterators (including the end iterators) or modifies any elements of the range involved, the behavior is undefined
Parameters
first last | The range of elements to fold. |
init | Initial value of the fold. |
op | Binary operation function object that will be applied. The signature of the function should be equivalent to the following:
|
Type requirements
InputIt | LegacyInputIterator |
T | CopyAssignable CopyConstructible |
Return value
acc
after all modifications.
Complexity
Exactly last - first
increments and assignments.
Exceptions
(none)
Possible implementation
accumulate(1)
template<class InputIt, class T>
constexpr // since C++20
T accumulate(InputIt first, InputIt last, T init)
{
for (; first != last; ++first)
init = std::move(init) + *first; // std::move since C++20
return init;
}
accumulate(2)
template<class InputIt, class T, class BinaryOperation>
constexpr // since C++20
T accumulate(InputIt first, InputIt last, T init, BinaryOperation op)
{
for (; first != last; ++first)
init = op(std::move(init), *first); // std::move since C++20
return init;
}
Notes
std::accumulate
performs a left fold. In order to perform a right fold, one must reverse the order of the arguments to the binary operator, and use reverse iterators.
If left to type inference, op
operates on values of the same type as init
which can result in unwanted casting of the iterator elements.
For example, std::accumulate(v.begin(), v.end(), 0)
likely does not give the result one wishes for when v
is of type std::vector<double>
.
Examples
#include <functional>
#include <iostream>
#include <numeric>
#include <string>
#include <vector>
int main()
{
std::vector<int> v {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int sum = std::accumulate(v.begin(), v.end(), 0);
int product = std::accumulate(v.begin(), v.end(), 1, std::multiplies<int>());
auto dash_fold = [](std::string a, int b)
{
return std::move(a) + '-' + std::to_string(b);
};
std::string s = std::accumulate(std::next(v.begin()), v.end(),
std::to_string(v[0]), // start with first element
dash_fold);
// Right fold using reverse iterators
std::string rs = std::accumulate(std::next(v.rbegin()), v.rend(),
std::to_string(v.back()), // start with last element
dash_fold);
std::cout << "sum: " << sum << '\n'
<< "product: " << product << '\n'
<< "dash-separated string: " << s << '\n'
<< "dash-separated string (right-folded): " << rs << '\n';
}
sum: 55
product: 3628800
dash-separated string: 1-2-3-4-5-6-7-8-9-10
dash-separated string (right-folded): 10-9-8-7-6-5-4-3-2-1
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