Range updation of values in c++ vectors [closed] - c++

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Given a vector of integers v, whose size is n and stores all 0's initially. If I want to update some elements of a range, say from index l to r, where (0<=l<r<n). Is there any efficient way instead of running loop from l to r?
For example,
v = {0,0,0,0,0,0,0}.
I want to update elements from l=2 to r=5.
Update refers to v[i] += k, where k is some known value. If k=1, then final updated vector in above case would be-
v = {0,0,1,1,1,1,0}.
Thanks:)

Answer 1

You can use for_each (ccpreference.com) or for_each_n
std::vector<int> data{0, 0, 0, 0, 0, 0, 0, 0};
std::for_each(data.begin() + 2, data.begin() + 6, [](int& i) { i += 1;});
// or with for_each_n
// std::for_each_n(data.begin() + 2, 4, [](int& i) { i += 1;});
for (const auto i : data) {
std::cout << i << ' ';
}
But as the comment above mentioned there will be a for loop in there, even though you don't explicitly write one.
Here a short explanation what the code does:
From the reference page we learn:
Applies the given function object f to the result of dereferencing every iterator in the range [first, last), in order.
In our case we provide for first data.begin() + 2 which is the 3rd element of the vector and for last we provide data.begin() + 6. This element is not included in the application of the function (this is what [first, last) means).
The last part is the function f that we have to provide. Here we use a lambda function that simply increases the parameter by 1. It's important that the function paramter of the lambda takes a reference (int& i) and not just a copy, otherwise the vector remain the same as we only change copies of the elements.