This is a text-only version of the following page on https://raymii.org:

This is a text-only version of the following page on https://raymii.org:
---
Title : std::accumulate in C++
Author : Remy van Elst
Date : 23-10-2019
Last update : 07-11-2020
URL : https://raymii.org/s/snippets/Cpp_std_accumulate.html
Format : Markdown/HTML
---

I'm using [codewars][1] to practice my development skills. Today I found out
about the `std::accumulate` method in C++ while doing an exercise there. I'm
sharing it here because I never heard of it before. It is the `<numeric>`
header, and it also accepts a custom binary function to apply instead of
`operator+`. This snippet shows some examples including a lambda operator and
the for loop you would use otherwise.

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Update 2020-11-08: Thank you Martin Blasko for reporting a few errors in this article!

### Plain old loop

It performs a [fold][2] on a given range. Or, in my case, it gives the
sum of all int's in a vector without a loop. [Complete documentation here][3].

Consider this code:

std::vector <int> numbers { 20, 10, -8, 10, 10, 15, 35 };
int sum = 0;
for (auto const& number: numbers)
{
sum += number;
}
std::cout << "sum: " << sum;

The output after compiling is:

# C:\Users\Remy\CLionProjects\codewars\cmake-build-debug\src\CodeWars.exe

sum: 92

This can be written shorter using the `accumulate` function:

int sum = std::accumulate(numbers.begin(), numbers.end(), 0);

`std::accumulate` accepts an initial value, so to add 10 to the result of the above sum without an
intermidiate variable:

std::vector <int> numbers { 20, 10, -8, 10, 10, 15, 35 };
int sum = std::accumulate(numbers.begin(), numbers.end(), 10);

The result will be `102`.

### Lambdas

`std::accumulate` also accepts a function to perform as the fold operator. For example, if you
need to get a sum of the result and also multiply every number by 8, in the
odd case you need to go from bytes to bits:

std::vector <int> numbers { 10, 20, 30 };
int sum = std::accumulate(numbers.begin(),
numbers.end(),
0,
[](int a, int b) { return a + (b*8);}
);
std::cout << "sum: " << sum;

Output:

sum: 480

The binary operator takes the current accumulation value a (initialized to init)
and the value of the current element b.

This is a simple example of course, but by using a lambda you can do all kinds
of crazy stuff.

From [the documentation][3], on reversing the order:

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.

[1]: https://www.codewars.com/r/KjbvJA
[2]: https://en.wikipedia.org/wiki/Fold_(higher-order_function)
[3]: https://en.cppreference.com/w/cpp/algorithm/accumulate

---

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