How can I get "address" of every Nth element of a vector without using "raw loop". std::copy_if seems to be a solution but I couldn't get the address of the element, I just could get the value.
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How do you want to "get the address"? Do you want to have a vector of the addresses? Call a function with each of the addresses? Have a function that produces each address? Have an input iterator that produces the addresses? – Yakk - Adam Nevraumont Feb 28 '19 at 18:33
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Sounds like you want a [skipping iterator](https://stackoverflow.com/questions/5685983/skipping-iterator) – NathanOliver Feb 28 '19 at 18:35
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vector of te addresses. I tried to std::transform but it returns all elements and I don't want to hold a counter and if statement in lambda. – zontragon Feb 28 '19 at 18:36
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2Vectors are stored in contiguous memory ranges. Therefore if you have the address of the 1st element `&v[0]`, you can simply calculate the m-times.n-th element as `&v[0] + m * n * sizeof
` where `T` is the type of the vector elements. – Frank Puffer Feb 28 '19 at 18:42
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You can use a skipping iterator (also called strided iterator) with std::for_each (or similar algorithm). The standard library has no generic skipping iterators however.
In case you wish to implement a skipping iterator yourself, the idea is simple: Write a template that takes an iterator as an argument. This template is an iterator "adaptor". Delegate most of the iterator boilerplate to the adapted iterator, but implement a custom increment operator, which skips elements based on your criteria.
While iterator adaptors are reusable, and existing implementations are available, a simple loop can be much less work:
for(size_t i = 0; i < v.size(); i += N)
eerorika
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