std::accumulate
(on partitioned ranges)
<numeric>
Computes the sum of the given value init and the elements in the range [
first,
last)
.
T
) with the initial value init and then modifies it with acc = acc + *i(until C++20)acc = std::move(acc) + *i(since C++20) for every iterator i in the range [
first,
last)
in order.T
) with the initial value init and then modifies it with acc = op(acc, *i)(until C++20)acc = op(std::move(acc), *i)(since C++20) for every iterator i in the range [
first,
last)
in order.If any of the following conditions is satisfied, the behavior is undefined:
T
is not CopyConstructible.
T
is not CopyAssignable.
[
first,
last)
.
[
first,
last]
.
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 type Type1 must be such that an object of type T can be implicitly converted to Type1. The type Type2 must be such that an object of type InputIt can be dereferenced and then implicitly converted to Type2. The type Ret must be such that an object of type T can be assigned a value of type Ret.
InputIt
must meet the requirements of LegacyInputIterator.
acc after all modifications.
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; } |
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>.
#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'; }
Output:
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
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
DR | Applied to | Behavior as published | Correct behavior |
---|---|---|---|
LWG 242 | C++98 | op could not have side effects | it cannot modify the ranges involved |