I'm basically looping through all the entries to check whether some entries is to be erased, but seems in a wrong way:
std::vector<HANDLE> myvector;
for(unsigned int i = 0; i < myvector.size(); i++)
{
if(...)
myvector.erase(myvector.begin()+i);
}
Anyone spot the problem in it? How to do it correctly?
You can use std::remove_if
. This will move all remaining elements to the front, and returns an iterator to the new back. You can then erase it:
struct my_predicate
{
bool operator()(HANDLE) const
{
return ...;
}
};
typedef std::vector<HANDLE> vector_type;
vector_type::iterator newEnd =
std::remove_if(myvector.begin(), myvector.end(), my_predicate());
myvector.erase(newEnd, myvector.end());
It's commonly done on one line. If your compiler supports lambda's (C++0x), you can do:
vector_type::iterator newEnd =
std::remove_if(myvector.begin(), myvector.end(), [](HANDLE){ return ... });
myvector.erase(newEnd, myvector.end());
To keep the predicate local.
If you think this is ugly, just wrap it up:
template <typename Vec, typename Pred>
Pred erase_if(Vec& pVec, Pred pPred)
{
pVec.erase(std::remove_if(pVec.begin(), pVec.end(),
pPred), pVec.end());
return pPred;
}
Then:
erase_if(myvector, mypredicate);
C++0x lambda's work the same, of course.
Your problem is algorithmic. What happens if two adjacent elements meet your criterion for deletion? The first will be deleted, but because i
is incremented after each iteration of the loop, the second will be skipped. This is because a vector is contiguous in memory, and all elements after the deleted one are moved forwards one index.
An ugly hack would be to do the following:
std::vector<HANDLE> myvector;
for(unsigned int i = 0; i < myvector.size();)
{
if(...)
myvector.erase(myvector.begin()+i);
else
i++;
}
I'm not sure if using iterators would work, because calling erase
invalidates iterators to elements after the erased element.
The elegant solution would be to use std::remove_if, as GMan suggested. This would abstract away two things:
Edit: I should also add, the hacked solution is O(n2) in the worst case. GMan's solution is O(n), assuming your removal condition is O(1). I would strongly encourage you to learn and use GMan's solution.
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