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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