Using std::accumulate on a two-dimensional std::array

Question

Given the two-dimensional array

std::array<std::array<int, 2>, 3> m = {{ {1, 2}, {3, 4}, {5, 6} }};

I am looking for the sum of all its elements - in this case, 21. Had the array been one-dimensional, I could've written

auto sum = std::accumulate(m.begin(), m.end(), 0);

but for my two-dimensional array, this fails with the rather understandable error

no match for 'operator+' (operand types are 'int' and 'std::array<int, 2ul>')

How can I elegantly compute this sum for my 2D array (avoiding for-loops, preferring STL-algorithms)?

Can it be done with a one-liner like for the one-dimensional case, or does it become more complex?

I don't have the energy to work out the details, but you can write an iterator that knows how to walk through a 2-dimensional array. Basically, it would walk through one row of the array and when it hits the end of the row, move to the next row. That's a little trickier to write than Rakete111's nested calls to std::accumulate, but it is more general: you can use that iterator for any algorithm. (If someone wants to work out the details, feel free to post your code as an answer) — Pete Becker
– Pete Becker, Commented Jun 28, 2017 at 17:11

Rakete1111 · Accepted Answer · 2017-06-30 18:56:39Z

23

It's just a bit more complex. You have to nest 2 std::accumulate calls. The nested std::accumulate call sums the elements in the nested arrays, and then the first std::accumulate sums those up.

auto sum = std::accumulate(m.cbegin(), m.cend(), 0, [](auto lhs, const auto& rhs) {
    return std::accumulate(rhs.cbegin(), rhs.cend(), lhs);
});

That's a C++14 solution because of the generic lambda, but for C++11, you just need to specify the types explicitly.

edited Jun 30, 2017 at 18:56

answered Jun 28, 2017 at 15:57

Rakete1111

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Serikov · Accepted Answer · 2017-06-29 13:35:55Z

10

Conceptually you want to flatten array m and then apply accumulate to it.
Using Range-v3 library (or Ranges TS in the future) you can do just that (link to wandbox).

std::array<std::array<int, 2>, 3> m = {{ {1, 2}, {3, 4}, {5, 6} }};

auto result = ranges::accumulate(ranges::join(m), 0); // flatten range then apply accumulate

That works like what Pete Becker mentioned in comment: "walk through one row of the array and when it hits the end of the row, move to the next row". No copy of subranges made.

edited Jun 29, 2017 at 13:35

answered Jun 29, 2017 at 13:10

Serikov

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1 Comment

Bart Vandewoestyne Over a year ago

Nice, I'm starting to like Ranges :-) Unfortunately, we're currently still stuck with VS2012, so only a few C++11 features available :-(

Aditya Kumar · Accepted Answer · 2021-07-20 12:54:44Z

1

Please try as below :

std::unique_ptr<int[][Column]> matrix (new int[Row][Column]);
long long total = 0;
for(int i=0; i < Row; i++){
total =  std::accumulate(std::begin(matrix[i]),std::end(matrix[i]),total);
}

answered Jul 20, 2021 at 12:54

Aditya Kumar

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