Comments (6)
The second parameter to the sort function is an beyond the end iterator, so the off by one "access" in this case intentional. However, a more correct solution here, and in a couple of other places in this lib, would be indices->begin() + node.last
. I am not sure when I'll have time to change that so I would be happy for a PR if you have the time ;-)
from libacc.
The problem is, that indices[node.last]
is no iterator here, but a std::vector access, so it throws an exception. As far as I understand the code, you're accessing the last+1 element and then take the address (which probably is a pointer to the underlying array).
I will have look if indices.begin() + node.first, indices.begin()+node.last
is the correct range for the iterator.
from libacc.
I know that indices[node.last] would be an out of bounds access, but depending on the compiler you don't need to access an element to take its address. But after writing this I have to agree that the code is broken considering that it relies on a compiler "optimization".
from libacc.
I need to use MSVC here and it breaks.
But looking at it, it seems that std::sort uses the points like an iterator, because both "*it" and "it++" are working on pointers as on iterators, but probably it is not meant to work on something else than iterators.
from libacc.
A pointer is an iterator, it is even a RandomAccessIterator since iterators are just concepts and a pointer implements all necessary functions to be classified as such.
A RandomAccessIterator is a BidirectionalIterator that can be moved to point to any element in constant time.
A pointer to an element of an array satisfies all requirements of RandomAccessIterator
http://en.cppreference.com/w/cpp/concept/RandomAccessIterator
Apart from that &vec[idx]
is incorrect for (idx >= vec.size()) because it can be broken down to &(vec[idx])
. I guess it never was an issue for me because the gcc stl has not bounds check on the array access operator and I never use the value of vec[idx]
so the address of was optimized.
from libacc.
Fixed in #4
from libacc.
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from libacc.